Samacheer Kalvi Class 10 Maths Solutions Chapter 6 Trigonometry More Ques

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions

I. Multiple Choice Questions

Question 1. (1 – sin² \( \theta \)) sec² \( \theta \) = ........
(a) 0
(b) 1
(c) tan² \( \theta \)
(d) cos \( \theta \)
Answer: (b) 1
Hint: We know that \( 1 - \sin^2 \theta = \cos^2 \theta \). Also, \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \). So, \( (1 - \sin^2 \theta) \sec^2 \theta = \cos^2 \theta \cdot \frac{1}{\cos^2 \theta} = 1 \). This means the expression simplifies to 1.
In simple words: The expression simplifies to 1 because \( 1 - \sin^2 \theta \) is the same as \( \cos^2 \theta \), and \( \sec^2 \theta \) is its reciprocal.

🎯 Exam Tip: Remember the fundamental trigonometric identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) and \( \sec^2 \theta - \tan^2 \theta = 1 \).

Question 2. (1 + tan² \( \theta \)) sin² \( \theta \) = ........
(a) sin² \( \theta \)
(b) cos² \( \theta \)
(c) tan² \( \theta \)
(d) cot² \( \theta \)
Answer: (c) tan² \( \theta \)
Hint: We know that \( 1 + \tan^2 \theta = \sec^2 \theta \). Also, \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \). So, \( (1 + \tan^2 \theta) \sin^2 \theta = \sec^2 \theta \sin^2 \theta = \frac{1}{\cos^2 \theta} \sin^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta \). This is a common identity simplification.
In simple words: When you combine \( 1 + \tan^2 \theta \) with \( \sin^2 \theta \), it simplifies to \( \tan^2 \theta \) because \( 1 + \tan^2 \theta \) is \( \sec^2 \theta \), which is \( \frac{1}{\cos^2 \theta} \).

🎯 Exam Tip: Always look for ways to simplify expressions using basic identities before performing further calculations.

Question 3. (1 − cos² \( \theta \)) (1 + cot² \( \theta \)) = ........
(a) sin² \( \theta \)
(b) 0
(c) 1
(d) tan² \( \theta \)
Answer: (c) 1
Hint: We know that \( 1 - \cos^2 \theta = \sin^2 \theta \) and \( 1 + \cot^2 \theta = \csc^2 \theta \). So, \( (1 - \cos^2 \theta) (1 + \cot^2 \theta) = \sin^2 \theta \cdot \csc^2 \theta = \sin^2 \theta \cdot \frac{1}{\sin^2 \theta} = 1 \). This simplification is very direct.
In simple words: The expression becomes 1 because the first part is \( \sin^2 \theta \) and the second part is \( \csc^2 \theta \), which are reciprocals of each other.

🎯 Exam Tip: Recognizing pairs of reciprocal trigonometric functions like sine and cosecant helps in quick simplification.

Question 4. sin (90° – \( \theta \)) cos \( \theta \) + cos (90° – \( \theta \)) sin \( \theta \) = ........
(a) 1
(b) 0
(c) 2
(d) -1
Answer: (a) 1
Hint: Using the complementary angle identities, \( \sin(90^\circ - \theta) = \cos \theta \) and \( \cos(90^\circ - \theta) = \sin \theta \).
\( \implies \) So the expression becomes \( \cos \theta \cdot \cos \theta + \sin \theta \cdot \sin \theta = \cos^2 \theta + \sin^2 \theta \).
\( \implies \) We know that \( \cos^2 \theta + \sin^2 \theta = 1 \). This is a basic identity.
In simple words: This problem uses two main ideas: first, that sine and cosine swap when the angle is \( 90^\circ - \theta \), and second, that \( \sin^2 \theta + \cos^2 \theta \) always equals 1.

🎯 Exam Tip: Be fluent with complementary angle identities and the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \).

Question 5. 1 - \( \frac{\sin^2 \theta}{1+\cos \theta} \) = ........
(a) cos \( \theta \)
(b) tan \( \theta \)
(c) cot \( \theta \)
(d) cosec \( \theta \)
Answer: (a) cos \( \theta \)
Hint: Start by finding a common denominator for the expression.
\( 1 - \frac{\sin^2 \theta}{1+\cos \theta} = \frac{1+\cos \theta - \sin^2 \theta}{1+\cos \theta} \)
\( \implies \) We know that \( \sin^2 \theta = 1 - \cos^2 \theta \). Substitute this into the numerator.
\( \implies \) \( = \frac{1+\cos \theta - (1-\cos^2 \theta)}{1+\cos \theta} = \frac{1+\cos \theta - 1 + \cos^2 \theta}{1+\cos \theta} = \frac{\cos \theta + \cos^2 \theta}{1+\cos \theta} \)
\( \implies \) Factor out \( \cos \theta \) from the numerator: \( = \frac{\cos \theta (1+\cos \theta)}{1+\cos \theta} \)
\( \implies \) Cancel out \( (1+\cos \theta) \): \( = \cos \theta \). This is a step-by-step simplification.
In simple words: To solve this, first combine the terms by finding a common bottom part. Then, use the rule that \( \sin^2 \theta \) is the same as \( 1 - \cos^2 \theta \). After that, you can simplify and cancel out terms to get \( \cos \theta \).

🎯 Exam Tip: When dealing with fractions and trigonometric identities, finding a common denominator and using Pythagorean identities are key steps for simplification.

Question 6. cos⁴ x - sin⁴ x = ........
(a) 2 sin² x - 1
(b) 2 cos² x - 1
(c) 1 + 2 sin² x
(d) 1 - 2 cos² x
Answer: (b) 2 cos² x - 1
Hint: This expression is in the form of \( a^2 - b^2 \).
\( \cos^4 x - \sin^4 x = (\cos^2 x)^2 - (\sin^2 x)^2 \)
\( \implies \) Using the difference of squares formula \( a^2 - b^2 = (a-b)(a+b) \), we get:
\( = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) \)
\( \implies \) We know that \( \cos^2 x + \sin^2 x = 1 \). So, the expression becomes:
\( = (\cos^2 x - \sin^2 x) \cdot 1 = \cos^2 x - \sin^2 x \)
\( \implies \) Now, substitute \( \sin^2 x = 1 - \cos^2 x \):
\( = \cos^2 x - (1 - \cos^2 x) = \cos^2 x - 1 + \cos^2 x = 2 \cos^2 x - 1 \). This expression is also an identity for \( \cos(2x) \).
In simple words: This problem uses a special algebra trick called 'difference of squares' and then a basic trigonometry rule. First, break down \( \cos^4 x - \sin^4 x \) into two parts. Then, simplify it to get \( 2 \cos^2 x - 1 \).

🎯 Exam Tip: Recognizing algebraic identities like \( a^2 - b^2 \) in trigonometric expressions is very useful. Also, remember \( \cos^2 x + \sin^2 x = 1 \).

Question 7. If tan \( \theta = \frac { a }{ x } \), then the value of \( \frac{x}{\sqrt{a^{2}+x^{2}}} \) = ........
(a) cos \( \theta \)
(b) sin \( \theta \)
(c) cosec \( \theta \)
(d) sec \( \theta \)
Answer: (a) cos \( \theta \)
Hint: Draw a right-angled triangle where \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \).
Given \( \tan \theta = \frac{a}{x} \). Let the opposite side be 'a' and the adjacent side be 'x'.
Using the Pythagorean theorem, the hypotenuse \( = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2} = \sqrt{a^2 + x^2} \).
Now, we need to find \( \frac{x}{\sqrt{a^2 + x^2}} \).
We know that \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \).
From our triangle, \( \cos \theta = \frac{x}{\sqrt{a^2 + x^2}} \). This matches the expression we need to evaluate.
In simple words: Draw a right triangle. If \( \tan \theta = \text{opposite} / \text{adjacent} \), then use Pythagoras to find the hypotenuse. Then, use the definition of \( \cos \theta \) as \( \text{adjacent} / \text{hypotenuse} \) to find the answer.

🎯 Exam Tip: When given a trigonometric ratio, sketching a right-angled triangle and labeling its sides can help visualize and find other ratios or expressions.

Question 8. If x = a sec \( \theta \), y = b tan \( \theta \), then the value of \( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} \) = ........
(a) 1
(b) tan² \( \theta \)
(c) cosec² \( \theta \)
(d) None of the options
Answer: (a) 1
Hint: Substitute the given values of x and y into the expression.
Given \( x = a \sec \theta \) and \( y = b \tan \theta \).
So, \( \frac{x^2}{a^2} = \frac{(a \sec \theta)^2}{a^2} = \frac{a^2 \sec^2 \theta}{a^2} = \sec^2 \theta \).
And, \( \frac{y^2}{b^2} = \frac{(b \tan \theta)^2}{b^2} = \frac{b^2 \tan^2 \theta}{b^2} = \tan^2 \theta \).
Now, substitute these into the expression \( \frac{x^2}{a^2} - \frac{y^2}{b^2} \):
\( = \sec^2 \theta - \tan^2 \theta \).
\( \implies \) We know the identity \( \sec^2 \theta - \tan^2 \theta = 1 \). So the value is 1.
In simple words: First, put the given values of x and y into the equation. Simplify each part to get \( \sec^2 \theta \) and \( \tan^2 \theta \). Then, use the basic math rule that \( \sec^2 \theta - \tan^2 \theta \) always equals 1.

🎯 Exam Tip: Remember the Pythagorean identity \( \sec^2 \theta - \tan^2 \theta = 1 \). This identity frequently appears in simplification and proof problems.

Question 9. \( \frac{\sec \theta}{\cot \theta+\tan \theta} \) = ........
(a) cot \( \theta \)
(b) tan \( \theta \)
(c) sin \( \theta \)
(d) – cot \( \theta \)
Answer: (c) sin \( \theta \)
Hint: Convert all terms to sine and cosine.
\( \sec \theta = \frac{1}{\cos \theta} \)
\( \cot \theta = \frac{\cos \theta}{\sin \theta} \)
\( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
Substitute these into the expression:
\( \frac{\frac{1}{\cos \theta}}{\frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta}} \)
\( \implies \) Simplify the denominator by finding a common denominator:
\( \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} \)
\( \implies \) Now, substitute the simplified denominator back into the main expression:
\( \frac{\frac{1}{\cos \theta}}{\frac{1}{\sin \theta \cos \theta}} = \frac{1}{\cos \theta} \cdot (\sin \theta \cos \theta) = \sin \theta \). This is a common way to simplify complex fractions.
In simple words: The easiest way to solve this is to change all the 'sec', 'cot', and 'tan' into 'sin' and 'cos'. After that, you'll need to fix the bottom part of the fraction by making a common denominator. Finally, simplify everything to get \( \sin \theta \).

🎯 Exam Tip: When simplifying complex trigonometric expressions, converting all terms into sine and cosine is often the most effective first step.

Question 10. \( \frac{\sin (90^\circ - \theta) \sin \theta}{\tan \theta} + \frac{\cos (90^\circ - \theta) \cos \theta}{\cot \theta} \) = ........
(a) tan \( \theta \)
(b) 1
(c) -1
(d) sin \( \theta \)
Answer: (b) 1
Hint: Use complementary angle identities: \( \sin(90^\circ - \theta) = \cos \theta \) and \( \cos(90^\circ - \theta) = \sin \theta \).
Substitute these into the expression:
\( = \frac{\cos \theta \cdot \sin \theta}{\tan \theta} + \frac{\sin \theta \cdot \cos \theta}{\cot \theta} \)
\( \implies \) We know \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \cot \theta = \frac{\cos \theta}{\sin \theta} \). Substitute these in the denominators:
\( = \frac{\cos \theta \sin \theta}{\frac{\sin \theta}{\cos \theta}} + \frac{\sin \theta \cos \theta}{\frac{\cos \theta}{\sin \theta}} \)
\( \implies \) Simplify each term:
\( = (\cos \theta \sin \theta) \cdot \frac{\cos \theta}{\sin \theta} + (\sin \theta \cos \theta) \cdot \frac{\sin \theta}{\cos \theta} \)
\( = \cos^2 \theta + \sin^2 \theta \)
\( \implies \) Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \). So the value is 1.
In simple words: First, swap sine and cosine for the \( 90^\circ - \theta \) angles. Then, change tan and cot into sine and cosine fractions. After simplifying the fractions, you will get \( \cos^2 \theta + \sin^2 \theta \), which always equals 1.

🎯 Exam Tip: Always prioritize converting expressions to sine and cosine and using complementary angle identities to simplify complicated trigonometric terms.

Question 11. In the adjoining figure, AC = ........
(a) 25m
(b) \(25 \sqrt{3}\) m
(c) \( \frac{25}{\sqrt{3}} \)
(d) \(25 \sqrt{2}\) m
Answer: (b) \(25 \sqrt{3}\) m
Hint: In the right-angled triangle ABC, with angle B = 90° and angle A = 60°, we want to find AC (the side opposite to angle B, which is BC in the image, but referred to as AC in the formula given).
We are given AB = 25m.
Using the tangent function: \( \tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB} \).
So, \( \tan 60^\circ = \frac{BC}{25} \).
\( \implies \sqrt{3} = \frac{BC}{25} \)
\( \implies BC = 25\sqrt{3} \) m. The problem labels the hypotenuse as AC, but the hint uses `AC` to mean the side opposite the angle. To match the answer, `AC` (in the hint's context) refers to the height `BC` in the diagram.
Therefore, the value is \( 25\sqrt{3} \) m. This calculates the vertical side based on the given base and angle.
In simple words: Look at the triangle. You know one side (25m) and one angle (60°). Use the 'tan' rule (opposite side divided by adjacent side) to find the length of the side labeled as AC in the hint, which is \( 25 \sqrt{3} \) meters.

🎯 Exam Tip: Always label the sides of a right-angled triangle (opposite, adjacent, hypotenuse) correctly with respect to the given angle to apply trigonometric ratios accurately.

Question 12. In the adjoining figure ∠ABC = ........
(a) 45°
(b) 30°
(c) 60°
(d) 50°
Answer: (c) 60°
Hint: In the right-angled triangle ABC, with angle C = 90°, we need to find angle B.
Given AC = \( 100\sqrt{3} \) m and BC = 100 m.
Using the tangent function for angle B: \( \tan B = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AC}{BC} \).
\( \implies \tan B = \frac{100\sqrt{3}}{100} = \sqrt{3} \).
\( \implies \) We know that \( \tan 60^\circ = \sqrt{3} \).
So, \( \angle B = 60^\circ \). This calculation directly gives the angle.
In simple words: Look at the triangle and the lengths of its sides. Use the 'tan' rule (opposite side divided by adjacent side) for angle B. Since the tan value is \( \sqrt{3} \), the angle B must be 60 degrees.

🎯 Exam Tip: Be familiar with the trigonometric values for common angles like 30°, 45°, and 60° to quickly solve such problems.

Question 13. A man is 28.5 m away from a tower. His eye level above the ground is 1.5 m. The angle of elevation of the tower from his eyes is 45°. Then the height of the tower is ........
(a) 30 m
(b) 27.5 m
(c) 28.5 m
(d) 27 m
Answer: (a) 30 m
Hint: Let AD be the total height of the tower. Let C be the man's eye level and E be his base. Let B be the point on the tower at the man's eye level.
From the figure, BC = 28.5 m (distance from man to tower).
DE = 1.5 m (man's eye level from ground). So, BD = CE = 1.5m.
In right-angled triangle ABC, \( \angle ACB = 45^\circ \).
We have \( \tan 45^\circ = \frac{AB}{BC} \).
\( \implies 1 = \frac{AB}{28.5} \)
\( \implies AB = 28.5 \) m. This is the height of the tower above the man's eye level.
The total height of the tower AD = AB + BD = 28.5 + 1.5 = 30 m. This is a common application of trigonometry.
In simple words: Draw a picture. Use the 'tan' rule with the 45° angle to find the height of the tower *above* the man's eyes. Then, add the man's eye level height to get the total height of the tower.

🎯 Exam Tip: When dealing with angle of elevation or depression problems involving eye-level, remember to add or subtract the observer's height from the calculated height to get the actual total height.

Question 14. In the adjoining figure, sin \( \theta = \frac { 15 }{ 17 } \) Then BC = ........
(a) 85 m
(b) 65 m
(c) 95 m
(d) 75 m
Answer: (d) 75 m
Hint: In the right-angled triangle ABC, with angle B = 90°, we are given \( \sin \theta = \frac{15}{17} \) where \( \theta \) is angle A.
We know that \( \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} \).
From the figure, AC = 85 m.
So, \( \frac{BC}{85} = \frac{15}{17} \).
\( \implies BC = \frac{15 \times 85}{17} \)
\( \implies BC = 15 \times 5 = 75 \) m. This directly calculates the length of side BC.
In simple words: In the right triangle, you know the angle (theta), its 'sin' value, and the longest side (hypotenuse). Use the 'sin' rule (opposite side divided by hypotenuse) to find the length of the side called BC.

🎯 Exam Tip: When using trigonometric ratios, ensure you correctly identify the opposite, adjacent, and hypotenuse sides relative to the given angle.

Question 15. (1 + tan² \( \theta \)) (1 − sin \( \theta \)) (1 + sin \( \theta \)) = ........
(a) cos² \( \theta \) – sin² \( \theta \)
(b) sin² \( \theta \) – cos² \( \theta \)
(c) sin² \( \theta \) + cos² \( \theta \)
(d) 0
Answer: (c) sin² \( \theta \) + cos² \( \theta \)
Hint: Start by simplifying the product of the last two terms.
We know that \( (a-b)(a+b) = a^2 - b^2 \). So, \( (1 - \sin \theta)(1 + \sin \theta) = 1^2 - \sin^2 \theta = 1 - \sin^2 \theta \).
We also know that \( 1 - \sin^2 \theta = \cos^2 \theta \).
The first term is \( (1 + \tan^2 \theta) \). We know that \( 1 + \tan^2 \theta = \sec^2 \theta \).
Now, substitute these simplified terms back into the original expression:
\( (1 + \tan^2 \theta) (1 - \sin \theta) (1 + \sin \theta) = (\sec^2 \theta) (\cos^2 \theta) \)
\( \implies \) Since \( \sec \theta = \frac{1}{\cos \theta} \), then \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \).
\( = \frac{1}{\cos^2 \theta} \cdot \cos^2 \theta = 1 \).
\( \implies \) And we know that \( \sin^2 \theta + \cos^2 \theta = 1 \). So, the answer is equivalent to \( \sin^2 \theta + \cos^2 \theta \). This uses multiple identities.
In simple words: First, combine the parts \( (1 - \sin \theta)(1 + \sin \theta) \) using a common algebra rule. Then, replace \( (1 + \tan^2 \theta) \) with its equivalent. Multiply the simplified parts. The final answer will be 1, which is the same as \( \sin^2 \theta + \cos^2 \theta \).

🎯 Exam Tip: Simplify expressions in stages, starting with easily recognizable algebraic or trigonometric identities, and always remember \( \sin^2 \theta + \cos^2 \theta = 1 \).

Question 16. (1 + cot² \( \theta \)) (1 – cos \( \theta \)) (1 + cos \( \theta \)) = ........
(a) tan² \( \theta \) – sec² \( \theta \)
(b) sin² \( \theta \) – cos² \( \theta \)
(c) sec² \( \theta \) – tan² \( \theta \)
(d) cos² \( \theta \) – sin² \( \theta \)
Answer: (c) sec² \( \theta \) – tan² \( \theta \)
Hint: Simplify the product of the last two terms first.
We know that \( (a-b)(a+b) = a^2 - b^2 \). So, \( (1 - \cos \theta)(1 + \cos \theta) = 1^2 - \cos^2 \theta = 1 - \cos^2 \theta \).
We also know that \( 1 - \cos^2 \theta = \sin^2 \theta \).
The first term is \( (1 + \cot^2 \theta) \). We know that \( 1 + \cot^2 \theta = \csc^2 \theta \).
Now, substitute these simplified terms back into the original expression:
\( (1 + \cot^2 \theta) (1 - \cos \theta) (1 + \cos \theta) = (\csc^2 \theta) (\sin^2 \theta) \)
\( \implies \) Since \( \csc \theta = \frac{1}{\sin \theta} \), then \( \csc^2 \theta = \frac{1}{\sin^2 \theta} \).
\( = \frac{1}{\sin^2 \theta} \cdot \sin^2 \theta = 1 \).
\( \implies \) And we know that \( \sec^2 \theta - \tan^2 \theta = 1 \). So, the answer is equivalent to \( \sec^2 \theta - \tan^2 \theta \). This is another common identity.
In simple words: First, simplify the parts \( (1 - \cos \theta)(1 + \cos \theta) \) using a basic algebra rule. Then, replace \( (1 + \cot^2 \theta) \) with its equivalent form. Multiply these simplified parts. The final answer will be 1, which is the same as \( \sec^2 \theta - \tan^2 \theta \).

🎯 Exam Tip: Practice recognizing and applying both algebraic identities and all three Pythagorean trigonometric identities to simplify expressions efficiently.

Question 17. (cos² \( \theta \) – 1) (cot² \( \theta \) + 1) + 1 = ........
(a) 1
(b) -1
(c) 2
(d) 0
Answer: (d) 0
Hint: Simplify each part of the expression using identities.
We know that \( \cos^2 \theta - 1 = - (1 - \cos^2 \theta) = - \sin^2 \theta \).
We also know that \( \cot^2 \theta + 1 = \csc^2 \theta \).
Now, substitute these into the expression:
\( (\cos^2 \theta - 1) (\cot^2 \theta + 1) + 1 = (-\sin^2 \theta) (\csc^2 \theta) + 1 \)
\( \implies \) Since \( \csc \theta = \frac{1}{\sin \theta} \), then \( \csc^2 \theta = \frac{1}{\sin^2 \theta} \).
\( = (-\sin^2 \theta) \cdot \frac{1}{\sin^2 \theta} + 1 \)
\( = -1 + 1 = 0 \). The expression simplifies completely.
In simple words: Change the first part using the rule that \( \cos^2 \theta - 1 \) is the same as \( -\sin^2 \theta \). Change the second part to \( \csc^2 \theta \). Then, multiply them and add 1. Everything will cancel out to 0.

🎯 Exam Tip: Pay close attention to negative signs when rearranging identities, such as \( \cos^2 \theta - 1 = -\sin^2 \theta \).

Question 18. \( \frac{1+\tan ^{2} \theta}{1+\cot ^{2} \theta} \) = ........
(a) cos² \( \theta \)
(b) tan² \( \theta \)
(c) sin² \( \theta \)
(d) None of the options
Answer: (b) tan² \( \theta \)
Hint: Use the Pythagorean identities for the numerator and denominator.
We know that \( 1 + \tan^2 \theta = \sec^2 \theta \).
We also know that \( 1 + \cot^2 \theta = \csc^2 \theta \).
Substitute these into the expression:
\( \frac{1+\tan^2 \theta}{1+\cot^2 \theta} = \frac{\sec^2 \theta}{\csc^2 \theta} \)
\( \implies \) Now, convert \( \sec^2 \theta \) and \( \csc^2 \theta \) into sine and cosine terms:
\( \sec^2 \theta = \frac{1}{\cos^2 \theta} \)
\( \csc^2 \theta = \frac{1}{\sin^2 \theta} \)
\( = \frac{\frac{1}{\cos^2 \theta}}{\frac{1}{\sin^2 \theta}} = \frac{1}{\cos^2 \theta} \cdot \sin^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \)
\( \implies \) We know that \( \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta \). This is a direct application of identities.
In simple words: Change the top part of the fraction to \( \sec^2 \theta \) and the bottom part to \( \csc^2 \theta \). Then, write \( \sec^2 \theta \) and \( \csc^2 \theta \) using \( \cos^2 \theta \) and \( \sin^2 \theta \). After simplifying, you will get \( \tan^2 \theta \).

🎯 Exam Tip: When faced with ratios of squared trigonometric functions, converting them to sine and cosine often reveals a simpler tangent or cotangent form.

Question 19. Sin² \( \theta \) + \( \frac{1}{1+\tan ^{2} \theta} \) = ........
(a) cosec² \( \theta \) + cot² \( \theta \)
(b) cosec² \( \theta \) – cot² \( \theta \)
(c) cot² \( \theta \) – cosec² \( \theta \)
(d) sin² \( \theta \) – cos² \( \theta \)
Answer: (b) cosec² \( \theta \) – cot² \( \theta \)
Hint: Simplify the fractional term first.
We know that \( 1 + \tan^2 \theta = \sec^2 \theta \).
So, \( \frac{1}{1+\tan^2 \theta} = \frac{1}{\sec^2 \theta} \).
We also know that \( \frac{1}{\sec^2 \theta} = \cos^2 \theta \).
Now, substitute this back into the original expression:
\( \sin^2 \theta + \frac{1}{1+\tan^2 \theta} = \sin^2 \theta + \cos^2 \theta \)
\( \implies \) We know that \( \sin^2 \theta + \cos^2 \theta = 1 \).
\( \implies \) And we know that \( \csc^2 \theta - \cot^2 \theta = 1 \). So, the answer is equivalent to \( \csc^2 \theta - \cot^2 \theta \). This uses fundamental identities.
In simple words: First, change the bottom part of the fraction using \( 1+\tan^2 \theta = \sec^2 \theta \). Then, simplify the fraction to \( \cos^2 \theta \). The whole problem becomes \( \sin^2 \theta + \cos^2 \theta \), which equals 1. This is the same as \( \csc^2 \theta - \cot^2 \theta \).

🎯 Exam Tip: Always look for ways to simplify complex terms using fundamental identities before combining them with other terms.

Question 20. 9 tan² \( \theta \) – 9 sec² \( \theta \) = ........
(a) 1
(b) 0
(c) 9
(d) -9
Answer: (d) -9
Hint: Factor out the common term, which is 9.
\( 9 \tan^2 \theta - 9 \sec^2 \theta = 9(\tan^2 \theta - \sec^2 \theta) \)
\( \implies \) We know the identity \( \sec^2 \theta - \tan^2 \theta = 1 \).
So, \( \tan^2 \theta - \sec^2 \theta = -(\sec^2 \theta - \tan^2 \theta) = -1 \).
Substitute this value back into the expression:
\( = 9(-1) = -9 \). This is a direct application of the identity.
In simple words: Take out the common number 9 from both parts. You will be left with \( \tan^2 \theta - \sec^2 \theta \). Remember that \( \sec^2 \theta - \tan^2 \theta \) is 1, so \( \tan^2 \theta - \sec^2 \theta \) is -1. Multiply -1 by 9 to get -9.

🎯 Exam Tip: Always factor out common terms first. Also, be careful with negative signs when rearranging identities like \( \sec^2 \theta - \tan^2 \theta = 1 \) into \( \tan^2 \theta - \sec^2 \theta = -1 \).

Question 21. The length of shadow of a tower on the plane ground is \( \sqrt { 3 } \) times the height of the tower. The angle of elevation of sum is ........
(a) 45°
(b) 30°
(c) 60°
(d) 90°
Answer: (b) 30°
Hint: Let the height of the tower be AB = x.
The length of the shadow on the ground is BC = \( \sqrt{3} \)x.
Let \( \theta \) be the angle of elevation of the sun (angle C).
In the right-angled triangle ABC, with angle B = 90°, we use the tangent function:
\( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} \)
\( \implies \tan \theta = \frac{x}{\sqrt{3}x} = \frac{1}{\sqrt{3}} \).
\( \implies \) We know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
So, \( \theta = 30^\circ \). This directly gives the angle of elevation.
In simple words: Draw a picture of the tower and its shadow. Use 'x' for the tower's height and '\( \sqrt{3} \)x' for the shadow. Use the 'tan' rule. When \( \tan \theta \) is \( 1/\sqrt{3} \), the angle is 30 degrees.

🎯 Exam Tip: Always draw a diagram for height and distance problems and correctly label the knowns and unknowns before applying trigonometric ratios.

Question 22. A ladder makes an angle of 60° with the ground, when placed against a wall. If the foot of the ladder is 2m away from the wall, then the length of the ladder (in metres) is ........
(a) \( \frac{4}{\sqrt{2}} \) m
(b) \( 4 \sqrt{3} \)m
(c) \( 2 \sqrt{2} \)m
(d) 4 m
Answer: (d) 4 m
Hint: Let AC be the length of the ladder, denoted by 'x'.
Let AB be the wall and BC be the ground. So, \( \triangle ABC \) is a right-angled triangle with angle B = 90°.
The ladder makes an angle of 60° with the ground, so \( \angle C = 60^\circ \).
The foot of the ladder is 2m away from the wall, so BC = 2m.
We need to find the length of the ladder (hypotenuse AC).
Using the cosine function: \( \cos C = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} \).
\( \implies \cos 60^\circ = \frac{2}{x} \)
\( \implies \frac{1}{2} = \frac{2}{x} \)
\( \implies x = 2 \times 2 = 4 \) m. The ladder is 4 meters long.
In simple words: Draw a picture of the ladder against the wall, forming a right triangle. You know the distance from the wall (adjacent side) and the angle (60°). Use the 'cos' rule (adjacent side divided by hypotenuse) to find the length of the ladder.

🎯 Exam Tip: For ladder problems, remember that the ladder itself is the hypotenuse, and the wall and ground form the other two sides of a right-angled triangle.

Question 23. The angle of depression of a car parked on the road from the top of a 150m tower is 30°. The distance of the car from the tower (in metres) is ........
(a) \( 150 \sqrt{3} \) m
(b) \( 150 \sqrt{2} \)
(c) 75 cm
(d) \( 50 \sqrt{3} \) m
Answer: (a) \( 150 \sqrt{3} \) m
Hint: Let AB be the height of the tower (150m) and C be the position of the car. Let BC be the distance of the car from the tower, denoted by 'x'.
The angle of depression from the top of the tower (A) to the car (C) is 30°.
This means the angle of elevation from the car (C) to the top of the tower (A) is also 30°, i.e., \( \angle ACB = 30^\circ \).
In the right-angled triangle ABC (angle B = 90°), we use the tangent function:
\( \tan 30^\circ = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} \)
\( \implies \frac{1}{\sqrt{3}} = \frac{150}{x} \)
\( \implies x = 150\sqrt{3} \) m. This is a standard calculation for this type of problem.
In simple words: Draw a triangle where the tower is the height and the car's distance is the base. The angle of depression from the top is the same as the angle of elevation from the bottom (30°). Use the 'tan' rule with the tower height and the angle to find the distance 'x'.

🎯 Exam Tip: For problems involving angles of depression, remember that the angle of depression from point A to point B is equal to the angle of elevation from point B to point A (alternate interior angles).

Question 24. The angle of elevation of the top of a tower at a point on the ground 50 m away from the foot of the tower is 45°. Then the height of the tower (in meters) is ........
(a) \( 50 \sqrt{3} \) m
(b) 50 m
(c) \( \frac{50}{\sqrt{2}} \) m
(d) \( \frac{50}{\sqrt{3}} \) m
Answer: (b) 50 m
Hint: Let AB be the height of the tower, denoted by 'x'. Let C be the point on the ground 50m away from the foot of the tower (B). So BC = 50m.
The angle of elevation of the top of the tower (A) from point C is 45°, so \( \angle ACB = 45^\circ \).
In the right-angled triangle ABC (angle B = 90°), we use the tangent function:
\( \tan 45^\circ = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} \)
\( \implies 1 = \frac{x}{50} \)
\( \implies x = 50 \) m. The height of the tower is 50 meters.
In simple words: Draw a triangle. The height of the tower is 'x', and the distance from the base is 50m. Since the angle of elevation is 45°, the height and distance are equal. So the tower is 50 meters tall.

🎯 Exam Tip: Remember that for an angle of elevation of 45°, the height of the object is equal to its horizontal distance from the observation point.

Question 25. If x = a cos \( \theta \) and y = b sin \( \theta \), then b²x² + a²y² = ........
(a) a²b²
(b) ab
(c) a⁴b⁴
(d) a² + b²
Answer: (a) a²b²
Hint: Substitute the given values of x and y into the expression.
Given \( x = a \cos \theta \) and \( y = b \sin \theta \).
Substitute these into \( b^2x^2 + a^2y^2 \):
\( = b^2(a \cos \theta)^2 + a^2(b \sin \theta)^2 \)
\( = b^2(a^2 \cos^2 \theta) + a^2(b^2 \sin^2 \theta) \)
\( = a^2b^2 \cos^2 \theta + a^2b^2 \sin^2 \theta \)
\( \implies \) Factor out \( a^2b^2 \):
\( = a^2b^2 (\cos^2 \theta + \sin^2 \theta) \)
\( \implies \) We know that \( \cos^2 \theta + \sin^2 \theta = 1 \).
\( = a^2b^2 \cdot 1 = a^2b^2 \). This uses a key trigonometric identity.
In simple words: Replace 'x' and 'y' in the equation with what they are equal to. Then, simplify the terms. You will see \( a^2b^2 \) appear in both parts. Take it out, and the rest will simplify to 1, leaving \( a^2b^2 \) as the answer.

🎯 Exam Tip: This type of problem often tests your ability to use the fundamental identity \( \sin^2 \theta + \cos^2 \theta = 1 \) after algebraic substitution.

II. Answer The Following Questions.

Question 1. Prove that sec² \( \theta \) + cosec² \( \theta \) = sec² \( \theta \) cosec² \( \theta \)
Answer: We need to prove that the left-hand side (L.H.S.) equals the right-hand side (R.H.S.).
L.H.S. \( = \sec^2 \theta + \csc^2 \theta \)
\( \implies \) Convert \( \sec^2 \theta \) and \( \csc^2 \theta \) to sine and cosine terms:
\( = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} \)
\( \implies \) Find a common denominator to add the fractions:
\( = \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta \sin^2 \theta} \)
\( \implies \) We know that \( \sin^2 \theta + \cos^2 \theta = 1 \). Substitute this into the numerator:
\( = \frac{1}{\cos^2 \theta \sin^2 \theta} \)
\( \implies \) Separate the terms and convert back to secant and cosecant:
\( = \frac{1}{\cos^2 \theta} \cdot \frac{1}{\sin^2 \theta} = \sec^2 \theta \csc^2 \theta \)
This is equal to the R.H.S. So, L.H.S. = R.H.S., and the identity is proven. This shows a link between addition and multiplication of these functions.
In simple words: Start by changing the left side to use only sine and cosine. Add the fractions by finding a common bottom part. Use the rule that \( \sin^2 \theta + \cos^2 \theta \) equals 1. Then, change back to secant and cosecant. You will see that it matches the right side of the equation.

🎯 Exam Tip: To prove identities involving sums or differences of reciprocal functions, convert them to sine and cosine first, then find a common denominator.

Question 2. Prove that \( \frac{\sin \theta}{1-\cos \theta} = \csc \theta + \cot \theta \).
Answer: We need to prove that the left-hand side (L.H.S.) equals the right-hand side (R.H.S.).
L.H.S. \( = \frac{\sin \theta}{1-\cos \theta} \)
\( \implies \) To simplify, multiply the numerator and denominator by the conjugate of the denominator, which is \( (1+\cos \theta) \):
\( = \frac{\sin \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta} \)
\( \implies \) Simplify the denominator using \( (a-b)(a+b) = a^2 - b^2 \):
\( = \frac{\sin \theta (1+\cos \theta)}{1^2 - \cos^2 \theta} = \frac{\sin \theta (1+\cos \theta)}{\sin^2 \theta} \)
\( \implies \) Cancel out one \( \sin \theta \) term from numerator and denominator:
\( = \frac{1+\cos \theta}{\sin \theta} \)
\( \implies \) Separate the fraction into two terms:
\( = \frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} \)
\( \implies \) Convert these terms to cosecant and cotangent:
\( = \csc \theta + \cot \theta \)
This is equal to the R.H.S. So, L.H.S. = R.H.S., and the identity is proven. This technique is often used when there is a \( (1 \pm \cos \theta) \) or \( (1 \pm \sin \theta) \) in the denominator.
In simple words: To prove this, take the left side. Multiply the top and bottom by \( (1+\cos \theta) \). This makes the bottom part \( \sin^2 \theta \). Then, you can cancel out a \( \sin \theta \) from the top and bottom. Split the remaining fraction into two parts, and you will see that it equals the right side.

🎯 Exam Tip: When the denominator is \( 1 \pm \sin \theta \) or \( 1 \pm \cos \theta \), multiplying by its conjugate is a very effective strategy to simplify the expression.

Question 3. Prove that \( \frac{\cos \theta}{\sec \theta-\tan \theta} = 1 + \sin \theta \).
Answer: We need to prove that the left-hand side (L.H.S.) equals the right-hand side (R.H.S.).
L.H.S. \( = \frac{\cos \theta}{\sec \theta-\tan \theta} \)
\( \implies \) Convert \( \sec \theta \) and \( \tan \theta \) to sine and cosine terms:
\( = \frac{\cos \theta}{\frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta}} \)
\( \implies \) Simplify the denominator by combining the fractions:
\( = \frac{\cos \theta}{\frac{1 - \sin \theta}{\cos \theta}} \)
\( \implies \) Invert the denominator and multiply:
\( = \cos \theta \cdot \frac{\cos \theta}{1 - \sin \theta} = \frac{\cos^2 \theta}{1 - \sin \theta} \)
\( \implies \) We know that \( \cos^2 \theta = 1 - \sin^2 \theta \). Substitute this into the numerator:
\( = \frac{1 - \sin^2 \theta}{1 - \sin \theta} \)
\( \implies \) Factor the numerator using \( a^2 - b^2 = (a-b)(a+b) \):
\( = \frac{(1 - \sin \theta)(1 + \sin \theta)}{1 - \sin \theta} \)
\( \implies \) Cancel out \( (1 - \sin \theta) \) from numerator and denominator:
\( = 1 + \sin \theta \)
This is equal to the R.H.S. So, L.H.S. = R.H.S., and the identity is proven. This involves transforming expressions into sine and cosine.
In simple words: Start with the left side. Change 'sec' and 'tan' into 'sin' and 'cos'. Simplify the bottom part into one fraction. Then, flip and multiply. Change \( \cos^2 \theta \) to \( 1 - \sin^2 \theta \), and use the \( a^2-b^2 \) rule to factor it. Cancel out matching parts, and you'll get the right side.

🎯 Exam Tip: When \( \sec \theta \pm \tan \theta \) or \( \csc \theta \pm \cot \theta \) appears in the denominator, converting to sine and cosine is a reliable strategy for simplification.

Question 4. Prove that sec \( \theta \) (1 – sin \( \theta \)) (sec \( \theta \) + tan \( \theta \)) = 1.
Answer: We need to prove that the left-hand side (L.H.S.) equals the right-hand side (R.H.S.).
L.H.S. \( = \sec \theta (1 - \sin \theta) (\sec \theta + \tan \theta) \)
\( \implies \) Convert all terms to sine and cosine:
\( = \frac{1}{\cos \theta} (1 - \sin \theta) \left(\frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}\right) \)
\( \implies \) Simplify the terms inside the parentheses:
\( = \frac{1}{\cos \theta} (1 - \sin \theta) \left(\frac{1 + \sin \theta}{\cos \theta}\right) \)
\( \implies \) Multiply the fractions:
\( = \frac{(1 - \sin \theta)(1 + \sin \theta)}{\cos \theta \cdot \cos \theta} \)
\( = \frac{1^2 - \sin^2 \theta}{\cos^2 \theta} \)
\( \implies \) We know that \( 1 - \sin^2 \theta = \cos^2 \theta \). Substitute this into the numerator:
\( = \frac{\cos^2 \theta}{\cos^2 \theta} \)
\( = 1 \)
This is equal to the R.H.S. So, L.H.S. = R.H.S., and the identity is proven. This involves converting to basic functions and using algebraic properties.
In simple words: First, change all 'sec' and 'tan' into 'sin' and 'cos'. Then, multiply the parts together, making sure to combine the two parts that look like \( (1-\sin \theta)(1+\sin \theta) \). Simplify the top and bottom of the fraction using the rule \( 1-\sin^2 \theta = \cos^2 \theta \), and you will get 1.

🎯 Exam Tip: When proving identities, converting all terms to sine and cosine is often the most systematic approach. Also, look for expressions that simplify using \( (a-b)(a+b) = a^2 - b^2 \).

Question 5. Prove that \( \frac{\sin \theta}{\csc \theta+\cot \theta}=1 - \cos \theta \).
Answer: We need to prove that the left-hand side (L.H.S.) equals the right-hand side (R.H.S.).
L.H.S. \( = \frac{\sin \theta}{\csc \theta+\cot \theta} \)
\( \implies \) Convert \( \csc \theta \) and \( \cot \theta \) to sine and cosine terms:
\( = \frac{\sin \theta}{\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta}} \)
\( \implies \) Simplify the denominator by combining the fractions:
\( = \frac{\sin \theta}{\frac{1 + \cos \theta}{\sin \theta}} \)
\( \implies \) Invert the denominator and multiply:
\( = \sin \theta \cdot \frac{\sin \theta}{1 + \cos \theta} = \frac{\sin^2 \theta}{1 + \cos \theta} \)
\( \implies \) We know that \( \sin^2 \theta = 1 - \cos^2 \theta \). Substitute this into the numerator:
\( = \frac{1 - \cos^2 \theta}{1 + \cos \theta} \)
\( \implies \) Factor the numerator using \( a^2 - b^2 = (a-b)(a+b) \):
\( = \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 + \cos \theta} \)
\( \implies \) Cancel out \( (1 + \cos \theta) \) from numerator and denominator:
\( = 1 - \cos \theta \)
This is equal to the R.H.S. So, L.H.S. = R.H.S., and the identity is proven. This shows a common way to simplify expressions with \( (1 \pm \cos \theta) \) terms.
In simple words: First, change 'csc' and 'cot' on the bottom into 'sin' and 'cos'. Combine the fractions on the bottom. Then, flip the bottom fraction and multiply it by the top. Change \( \sin^2 \theta \) to \( 1 - \cos^2 \theta \). Factor this using the \( a^2-b^2 \) rule, and then cancel parts to get \( 1 - \cos \theta \).

🎯 Exam Tip: When you see \( \sin^2 \theta \) or \( \cos^2 \theta \) in an expression, consider substituting \( 1-\cos^2 \theta \) or \( 1-\sin^2 \theta \) respectively, especially if you have \( (1 \pm \cos \theta) \) or \( (1 \pm \sin \theta) \) terms.

Question 6. Prove the identity \( \frac{\sin \theta}{\csc \theta}+\frac{\cos \theta}{\sec \theta}=1 \).
Answer: We need to prove that the left-hand side (L.H.S.) equals the right-hand side (R.H.S.).
L.H.S. \( = \frac{\sin \theta}{\csc \theta} + \frac{\cos \theta}{\sec \theta} \)
\( \implies \) Convert \( \csc \theta \) and \( \sec \theta \) to sine and cosine terms:
\( \csc \theta = \frac{1}{\sin \theta} \)
\( \sec \theta = \frac{1}{\cos \theta} \)
Substitute these into the expression:
\( = \frac{\sin \theta}{\frac{1}{\sin \theta}} + \frac{\cos \theta}{\frac{1}{\cos \theta}} \)
\( \implies \) Simplify each term by multiplying by the reciprocal of the denominator:
\( = \sin \theta \cdot \sin \theta + \cos \theta \cdot \cos \theta \)
\( = \sin^2 \theta + \cos^2 \theta \)
\( \implies \) We know that \( \sin^2 \theta + \cos^2 \theta = 1 \).
This is equal to the R.H.S. So, L.H.S. = R.H.S., and the identity is proven. This is a fundamental trigonometric identity.
In simple words: Change 'csc' to \( 1/\sin \theta \) and 'sec' to \( 1/\cos \theta \). Then, simplify each part of the sum. You will get \( \sin^2 \theta + \cos^2 \theta \), which always equals 1.

🎯 Exam Tip: Always remember that cosecant is the reciprocal of sine, and secant is the reciprocal of cosine. This helps simplify fractions quickly.

Question 7. Prove the identity \( \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} = \csc \theta – \cot \theta \).
Answer: We need to prove that the left-hand side (L.H.S.) equals the right-hand side (R.H.S.).
L.H.S. \( = \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \)
\( \implies \) To remove the square root and simplify the denominator, multiply the numerator and denominator inside the square root by \( (1-\cos \theta) \):
\( = \sqrt{\frac{(1-\cos \theta)(1-\cos \theta)}{(1+\cos \theta)(1-\cos \theta)}} \)
\( \implies \) Simplify the numerator and denominator:
\( = \sqrt{\frac{(1-\cos \theta)^2}{1^2 - \cos^2 \theta}} = \sqrt{\frac{(1-\cos \theta)^2}{\sin^2 \theta}} \)
\( \implies \) Take the square root of the numerator and denominator:
\( = \frac{\sqrt{(1-\cos \theta)^2}}{\sqrt{\sin^2 \theta}} = \frac{|1-\cos \theta|}{|\sin \theta|} \)
Since \( 1-\cos \theta \) is always non-negative and \( \sin \theta \) can be positive or negative, we assume \( \sin \theta > 0 \) for simplification.
\( = \frac{1-\cos \theta}{\sin \theta} \)
\( \implies \) Separate the fraction into two terms:
\( = \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} \)
\( \implies \) Convert these terms to cosecant and cotangent:
\( = \csc \theta - \cot \theta \)
This is equal to the R.H.S. So, L.H.S. = R.H.S., and the identity is proven. This technique is often used when square roots are involved.
In simple words: Start with the left side. Multiply the top and bottom inside the square root by \( (1-\cos \theta) \). This helps to simplify the bottom to \( \sin^2 \theta \). Take the square root of both top and bottom. Then, separate the fraction into two parts. You will get the right side, \( \csc \theta - \cot \theta \).

🎯 Exam Tip: When an expression involves \( \sqrt{\frac{1 \pm \cos \theta}{1 \mp \cos \theta}} \) or \( \sqrt{\frac{1 \pm \sin \theta}{1 \mp \sin \theta}} \), rationalize the denominator (inside the square root) by multiplying by the conjugate.

Question 8. Prove the identity [cosec (90° – \( \theta \)) – sin (90° – \( \theta \))] [cosec \( \theta \) – sin \( \theta \)] [tan \( \theta \) + cot \( \theta \)] = 1.
Answer: We need to prove that the left-hand side (L.H.S.) equals the right-hand side (R.H.S.).
L.H.S. \( = [\csc(90^\circ - \theta) - \sin(90^\circ - \theta)] [\csc \theta - \sin \theta] [\tan \theta + \cot \theta] \)
\( \implies \) Use complementary angle identities: \( \csc(90^\circ - \theta) = \sec \theta \) and \( \sin(90^\circ - \theta) = \cos \theta \).
So, the first bracket becomes \( [\sec \theta - \cos \theta] \).
\( \implies \) Now, convert all terms to sine and cosine:
First bracket: \( \left[\frac{1}{\cos \theta} - \cos \theta\right] = \left[\frac{1 - \cos^2 \theta}{\cos \theta}\right] = \left[\frac{\sin^2 \theta}{\cos \theta}\right] \)
Second bracket: \( \left[\frac{1}{\sin \theta} - \sin \theta\right] = \left[\frac{1 - \sin^2 \theta}{\sin \theta}\right] = \left[\frac{\cos^2 \theta}{\sin \theta}\right] \)
Third bracket: \( \left[\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}\right] = \left[\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}\right] = \left[\frac{1}{\sin \theta \cos \theta}\right] \)
\( \implies \) Multiply all three simplified brackets:
\( = \left(\frac{\sin^2 \theta}{\cos \theta}\right) \cdot \left(\frac{\cos^2 \theta}{\sin \theta}\right) \cdot \left(\frac{1}{\sin \theta \cos \theta}\right) \)
\( \implies \) Cancel common terms from numerator and denominator:
\( = \frac{\sin^2 \theta \cos^2 \theta}{\cos \theta \sin \theta \sin \theta \cos \theta} = \frac{\sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \)
\( = 1 \)
This is equal to the R.H.S. So, L.H.S. = R.H.S., and the identity is proven. This involves multiple steps of simplification.
In simple words: First, change the parts with \( 90^\circ - \theta \) to their equivalent forms. Then, change all 'sec', 'csc', 'tan', and 'cot' to 'sin' and 'cos'. Simplify each of the three brackets separately. Finally, multiply all three simplified results together. You will see that everything cancels out to 1.

🎯 Exam Tip: For complex identities with multiple factors, simplify each factor separately first, then multiply them to achieve the final result. Converting to sine and cosine is a powerful strategy.

Question 4. Prove that \( \sec \theta (1 – \sin \theta) (\sec \theta + \tan \theta) = 1 \)
Answer: We start with the left side of the equation.
L.H.S \( = \sec \theta (1 – \sin \theta) (\sec \theta + \tan \theta) \)
First, change secant and tangent into sine and cosine:
\( = \frac{1}{\cos \theta} (1 – \sin \theta) (\frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}) \)
Next, combine the terms inside the second parenthesis:
\( = \frac{1 – \sin \theta}{\cos \theta} \times \frac{1 + \sin \theta}{\cos \theta} \)
Now, multiply the numerators and denominators:
\( = \frac{(1 – \sin \theta)(1 + \sin \theta)}{\cos^2 \theta} \)
Using the identity \( (a-b)(a+b) = a^2-b^2 \), the numerator becomes:
\( = \frac{1 – \sin^2 \theta}{\cos^2 \theta} \)
We know that \( 1 – \sin^2 \theta = \cos^2 \theta \). Substitute this into the expression:
\( = \frac{\cos^2 \theta}{\cos^2 \theta} \)
\( = 1 \)
Thus, the left-hand side equals the right-hand side. This shows the identity is correct.
In simple words: We change everything to sines and cosines, then use simple algebra to combine the terms. Because \( 1 - \sin^2 \theta \) is the same as \( \cos^2 \theta \), the whole expression simplifies to 1.

🎯 Exam Tip: When proving trigonometric identities, converting all terms to sine and cosine is often a good first step, especially for complex expressions, as it makes the relationships clearer.

Question 5. Prove that \( \frac{\sin \theta}{\csc \theta+\cot \theta} = 1 – \cos \theta \)
Answer: We begin with the left side of the equation.
L.H.S \( = \frac{\sin \theta}{\csc \theta+\cot \theta} \)
Let's convert cosecant and cotangent into sine and cosine terms:
\( = \frac{\sin \theta}{(\frac{1}{\sin \theta}) + (\frac{\cos \theta}{\sin \theta})} \)
Combine the terms in the denominator:
\( = \frac{\sin \theta}{\frac{1 + \cos \theta}{\sin \theta}} \)
Now, multiply by the reciprocal of the denominator:
\( = \sin \theta \times \frac{\sin \theta}{1 + \cos \theta} \)
\( = \frac{\sin^2 \theta}{1 + \cos \theta} \)
We use the Pythagorean identity \( \sin^2 \theta = 1 – \cos^2 \theta \). Substitute this into the numerator:
\( = \frac{1 – \cos^2 \theta}{1 + \cos \theta} \)
Factor the numerator using \( a^2 – b^2 = (a-b)(a+b) \):
\( = \frac{(1 – \cos \theta)(1 + \cos \theta)}{1 + \cos \theta} \)
Cancel out the common term \( (1 + \cos \theta) \):
\( = 1 – \cos \theta \)
This is equal to the right-hand side, so the identity is proven.
In simple words: First, change all parts of the fraction to use only sine and cosine. Then simplify the fraction, and use the rule that \( \sin^2 \theta \) can be replaced by \( 1 - \cos^2 \theta \). This helps us cancel terms and get the answer.

🎯 Exam Tip: When dealing with fractions in trigonometric identities, try to simplify the numerator and denominator separately before dividing. Remember common identities like \( \sin^2 \theta + \cos^2 \theta = 1 \).

Question 6. Prove the identity \( \frac{\sin \theta}{\csc \theta}+\frac{\cos \theta}{\sec \theta} = 1 \)
Answer: Let's start with the left side of the identity.
L.H.S \( = \frac{\sin \theta}{\csc \theta}+\frac{\cos \theta}{\sec \theta} \)
We know that \( \csc \theta = \frac{1}{\sin \theta} \) and \( \sec \theta = \frac{1}{\cos \theta} \). Substitute these reciprocal identities:
\( = \frac{\sin \theta}{(\frac{1}{\sin \theta})} + \frac{\cos \theta}{(\frac{1}{\cos \theta})} \)
When you divide by a fraction, you multiply by its reciprocal:
\( = \sin \theta \times \sin \theta + \cos \theta \times \cos \theta \)
\( = \sin^2 \theta + \cos^2 \theta \)
According to the fundamental Pythagorean identity, \( \sin^2 \theta + \cos^2 \theta \) is always equal to 1.
\( = 1 \)
This matches the right-hand side of the equation, so the identity is proven.
In simple words: This problem uses simple rules about how sine and cosine are related to cosecant and secant. Once you replace those with sines and cosines, you get \( \sin^2 \theta + \cos^2 \theta \), which always equals 1.

🎯 Exam Tip: Remembering reciprocal identities (like \( \csc \theta = 1/\sin \theta \)) and the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) is crucial for solving many trigonometry proofs quickly.

Question 7. Prove the identity \( \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} = \csc \theta – \cot \theta \)
Answer: Let's begin with the left side of the identity.
L.H.S \( = \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \)
To simplify the square root, we multiply the numerator and denominator inside the square root by \( (1 – \cos \theta) \):
\( = \sqrt{\frac{(1 – \cos \theta)(1 – \cos \theta)}{(1 + \cos \theta)(1 – \cos \theta)}} \)
This gives us:
\( = \sqrt{\frac{(1 – \cos \theta)^2}{1 – \cos^2 \theta}} \)
Using the Pythagorean identity \( 1 – \cos^2 \theta = \sin^2 \theta \):
\( = \sqrt{\frac{(1 – \cos \theta)^2}{\sin^2 \theta}} \)
Now, we can take the square root of the numerator and the denominator:
\( = \frac{1 – \cos \theta}{\sin \theta} \)
Separate the fraction into two terms:
\( = \frac{1}{\sin \theta} – \frac{\cos \theta}{\sin \theta} \)
Finally, use the reciprocal identity \( \frac{1}{\sin \theta} = \csc \theta \) and the quotient identity \( \frac{\cos \theta}{\sin \theta} = \cot \theta \):
\( = \csc \theta – \cot \theta \)
This matches the right-hand side of the equation, so the identity is proven.
In simple words: To prove this, we multiply the top and bottom of the fraction under the square root by \( 1 - \cos \theta \). This helps us use the \( \sin^2 \theta + \cos^2 \theta = 1 \) rule. After taking the square root, we separate the fraction to get cosecant and cotangent.

🎯 Exam Tip: When you see expressions with \( (1 \pm \cos \theta) \) or \( (1 \pm \sin \theta) \) under a square root, multiplying by the conjugate is usually the most effective strategy to simplify the expression.

Question 8. Prove the identity \( [\csc (90^\circ – \theta) – \sin (90^\circ – \theta)] (\csc \theta – \sin \theta) (\tan \theta + \cot \theta) = 1 \)
Answer: Let's simplify the left-hand side of the identity.
L.H.S \( = [\csc (90^\circ – \theta) – \sin (90^\circ – \theta)] (\csc \theta – \sin \theta) (\tan \theta + \cot \theta) \)
Using the complementary angle identities, \( \csc (90^\circ – \theta) = \sec \theta \) and \( \sin (90^\circ – \theta) = \cos \theta \):
\( = (\sec \theta – \cos \theta) (\csc \theta – \sin \theta) (\tan \theta + \cot \theta) \)
Now, convert all terms to sine and cosine:
\( = \left(\frac{1}{\cos \theta} – \cos \theta\right) \left(\frac{1}{\sin \theta} – \sin \theta\right) \left(\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}\right) \)
Combine the terms within each parenthesis:
\( = \left(\frac{1 – \cos^2 \theta}{\cos \theta}\right) \left(\frac{1 – \sin^2 \theta}{\sin \theta}\right) \left(\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}\right) \)
Apply the Pythagorean identity \( 1 – \cos^2 \theta = \sin^2 \theta \) and \( 1 – \sin^2 \theta = \cos^2 \theta \), and \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = \left(\frac{\sin^2 \theta}{\cos \theta}\right) \left(\frac{\cos^2 \theta}{\sin \theta}\right) \left(\frac{1}{\sin \theta \cos \theta}\right) \)
Multiply all the fractions:
\( = \frac{\sin^2 \theta \cos^2 \theta \times 1}{\cos \theta \sin \theta \sin \theta \cos \theta} \)
Simplify by canceling out terms from the numerator and denominator:
\( = \frac{\sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \)
\( = 1 \)
This result matches the right-hand side of the identity, proving it.
In simple words: We first change the angle for cosecant and sine, then rewrite all terms using only sine and cosine. After combining fractions and using the \( \sin^2 \theta + \cos^2 \theta = 1 \) rule, many terms cancel out, leaving just 1.

🎯 Exam Tip: When you have products of multiple trigonometric expressions, convert them to sines and cosines first. This often reveals opportunities to simplify using Pythagorean identities and cancellations.

Question 9. A kite is flying with a string of length 200 m. If the thread makes an angle 30° with the ground, find the distance of the kite from the ground level. (Here, assume that the string is along a straight line.)
Answer: Let's imagine this situation as a right-angled triangle. The length of the string is the hypotenuse, and the height of the kite from the ground is the opposite side to the angle of elevation.
Let \( h \) be the height (distance) of the kite from the ground level.
The length of the string (hypotenuse) \( = 200 \) m.
The angle of elevation \( = 30^\circ \).
We can use the sine function, which relates the opposite side and the hypotenuse:
\( \sin (\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}} \)
\( \sin 30^\circ = \frac{h}{200} \)
We know that \( \sin 30^\circ = \frac{1}{2} \).
\( \frac{1}{2} = \frac{h}{200} \)
To find \( h \), multiply both sides by 200:
\( h = 200 \times \frac{1}{2} \)
\( h = 100 \) m.
So, the kite is 100 meters above the ground.
In simple words: The kite string, the ground, and the kite's height make a triangle. We know the string length and the angle. Using the sine rule, we found the height of the kite above the ground.

🎯 Exam Tip: Always draw a simple diagram for height and distance problems to visualize the right-angled triangle. Correctly identifying the hypotenuse, opposite, and adjacent sides with respect to the given angle is key.

Question 10. Find the angular elevation (angle of elevation from the ground level) of the Sun when the length of the shadow of a 30 m long pole is \( 10 \sqrt { 3 } \) m.
Answer: Let's set up this problem as a right-angled triangle. The pole stands vertically, forming the opposite side, and its shadow is on the ground, forming the adjacent side.
Height of the pole (opposite side) \( = 30 \) m.
Length of the shadow (adjacent side) \( = 10 \sqrt{3} \) m.
Let \( \theta \) be the angle of elevation of the Sun.
We use the tangent function, which relates the opposite side and the adjacent side:
\( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)
\( \tan \theta = \frac{30}{10\sqrt{3}} \)
Simplify the fraction:
\( \tan \theta = \frac{3}{\sqrt{3}} \)
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} \):
\( \tan \theta = \frac{3\sqrt{3}}{3} \)
\( \tan \theta = \sqrt{3} \)
We know that \( \tan 60^\circ = \sqrt{3} \).
Therefore, \( \theta = 60^\circ \).
The angular elevation of the Sun is \( 60^\circ \).
In simple words: We have a pole and its shadow, forming a right triangle. The pole's height is opposite the angle, and the shadow is adjacent. Using the tangent rule, we find the angle of the sun's elevation.

🎯 Exam Tip: Remember standard trigonometric values for common angles like \( 30^\circ, 45^\circ, 60^\circ \). Rationalizing denominators (like \( 3/\sqrt{3} \) to \( \sqrt{3} \)) helps in recognizing these values.

Question 11. A ramp for unloading a moving truck, has an angle of elevation of 30°. If the top of the ramp is 0.9 m above the ground level, then find the length of the ramp.
Answer: Let's visualize this scenario as a right-angled triangle. The ramp forms the hypotenuse, and the height of the ramp above the ground is the side opposite the angle of elevation.
Let \( x \) be the length of the ramp.
The height of the top of the ramp above the ground (opposite side) \( = 0.9 \) m.
The angle of elevation \( = 30^\circ \).
We use the sine function, which connects the opposite side and the hypotenuse:
\( \sin (\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}} \)
\( \sin 30^\circ = \frac{0.9}{x} \)
We know that \( \sin 30^\circ = \frac{1}{2} \).
\( \frac{1}{2} = \frac{0.9}{x} \)
To solve for \( x \), cross-multiply:
\( x \times 1 = 0.9 \times 2 \)
\( x = 1.8 \) m.
Thus, the length of the ramp is 1.8 meters.
In simple words: The ramp's height and length form a right triangle with the ground. Knowing the height and the angle, we use the sine rule to easily find the ramp's full length.

🎯 Exam Tip: Always make sure your units are consistent (e.g., all meters). For questions involving height, distance, and angles, drawing a simple sketch can help clarify which trigonometric ratio to use (sine, cosine, or tangent).

Question 12. A girl of height 150 cm stands in front of a lamp-post and casts a shadow of length \( 150 \sqrt { 3 } \) cm on the ground. Find the angle of elevation of the top of the lamp-post.
Answer: Let's represent this situation using a right-angled triangle. The girl's height acts as the perpendicular (opposite side), and her shadow on the ground acts as the base (adjacent side). The angle of elevation is the angle formed at the end of the shadow to the top of her head.
Height of the girl (opposite side) \( = 150 \) cm.
Length of the shadow (adjacent side) \( = 150 \sqrt{3} \) cm.
Let \( \theta \) be the angle of elevation.
We use the tangent function, which relates the opposite side and the adjacent side:
\( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)
\( \tan \theta = \frac{150}{150\sqrt{3}} \)
Simplify the fraction by canceling 150 from the numerator and denominator:
\( \tan \theta = \frac{1}{\sqrt{3}} \)
We know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
Therefore, \( \theta = 30^\circ \).
The angle of elevation of the top of the lamp-post (from the context of the girl's shadow) is \( 30^\circ \).
In simple words: The girl's height and her shadow create a right triangle. By dividing her height by her shadow length, we get \( \tan \theta \), which shows the angle of elevation is \( 30^\circ \).

🎯 Exam Tip: Always look for common trigonometric values like \( 1/\sqrt{3} \), \( 1 \), or \( \sqrt{3} \), which correspond to angles like \( 30^\circ, 45^\circ, 60^\circ \). This helps quickly identify the angle.

Question 13. Prove that \( \sqrt{\cot ^{2} \theta-\cos ^{2} \theta} = \cot \theta \cdot \cos \theta \)
Answer: Let's simplify the left-hand side of the identity.
L.H.S \( = \sqrt{\cot^2 \theta – \cos^2 \theta} \)
First, express \( \cot^2 \theta \) in terms of sine and cosine: \( \cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta} \).
\( = \sqrt{\frac{\cos^2 \theta}{\sin^2 \theta} – \cos^2 \theta} \)
Factor out \( \cos^2 \theta \) from both terms under the square root:
\( = \sqrt{\cos^2 \theta \left(\frac{1}{\sin^2 \theta} – 1\right)} \)
We know that \( \frac{1}{\sin^2 \theta} = \csc^2 \theta \). Substitute this in:
\( = \sqrt{\cos^2 \theta (\csc^2 \theta – 1)} \)
Use the identity \( \csc^2 \theta – 1 = \cot^2 \theta \):
\( = \sqrt{\cos^2 \theta \cot^2 \theta} \)
Since the terms are squared under the square root, we can take them out:
\( = |\cos \theta \cot \theta| \)
Assuming \( \theta \) is in a quadrant where \( \cos \theta \cot \theta \) is non-negative (for which the identity holds):
\( = \cos \theta \cot \theta \)
This matches the right-hand side of the identity, proving it.
In simple words: To prove this, we change cotangent to cosine over sine. Then, we factor out \( \cos^2 \theta \) and use the rule that \( \csc^2 \theta - 1 \) is \( \cot^2 \theta \). This simplifies the whole expression under the square root.

🎯 Exam Tip: When dealing with square roots in trigonometric identities, try to manipulate the expression inside the root to get perfect squares, often by factoring out common terms or using Pythagorean identities.

Question 14. Prove that \( \frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta} = 2 \sec \theta \)
Answer: Let's simplify the left-hand side of the identity.
L.H.S \( = \frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta} \)
To add these fractions, find a common denominator, which is \( \cos \theta (1+\sin \theta) \):
\( = \frac{(1+\sin \theta)(1+\sin \theta) + \cos \theta \cdot \cos \theta}{\cos \theta (1+\sin \theta)} \)
\( = \frac{(1+\sin \theta)^2 + \cos^2 \theta}{\cos \theta (1+\sin \theta)} \)
Expand \( (1+\sin \theta)^2 \) as \( 1 + 2\sin \theta + \sin^2 \theta \):
\( = \frac{1 + 2\sin \theta + \sin^2 \theta + \cos^2 \theta}{\cos \theta (1+\sin \theta)} \)
Use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = \frac{1 + 2\sin \theta + 1}{\cos \theta (1+\sin \theta)} \)
Combine the constant terms:
\( = \frac{2 + 2\sin \theta}{\cos \theta (1+\sin \theta)} \)
Factor out 2 from the numerator:
\( = \frac{2(1 + \sin \theta)}{\cos \theta (1+\sin \theta)} \)
Cancel the common term \( (1 + \sin \theta) \) from the numerator and denominator:
\( = \frac{2}{\cos \theta} \)
Since \( \frac{1}{\cos \theta} = \sec \theta \), we have:
\( = 2\sec \theta \)
This matches the right-hand side of the identity, so it is proven.
In simple words: We combine the two fractions by finding a common bottom part. Expanding the top and using the rule \( \sin^2 \theta + \cos^2 \theta = 1 \) helps simplify it a lot. Finally, we cancel common terms and rewrite it using secant.

🎯 Exam Tip: When adding or subtracting trigonometric fractions, always find a common denominator. Look for opportunities to expand terms and apply Pythagorean identities to simplify the numerator.

Question 15. A tower is \( 100 \sqrt { 3 } \) metres high. Find the angle of elevation of its top from a point 100 metres away from its foot.
Answer: Let's form a right-angled triangle to represent this situation. The tower's height is the side opposite the angle of elevation, and the distance from the tower's foot is the adjacent side.
Height of the tower (opposite side) \( = 100\sqrt{3} \) m.
Distance from the foot of the tower (adjacent side) \( = 100 \) m.
Let \( \theta \) be the angle of elevation.
We use the tangent function, which relates the opposite side and the adjacent side:
\( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)
\( \tan \theta = \frac{100\sqrt{3}}{100} \)
Cancel out 100 from the numerator and denominator:
\( \tan \theta = \sqrt{3} \)
We know that \( \tan 60^\circ = \sqrt{3} \).
Therefore, \( \theta = 60^\circ \).
The angle of elevation of the tower's top is \( 60^\circ \).
In simple words: We have a tower's height and the distance from its base, which form a right triangle. Using the tangent rule, we divide the height by the distance. Since the answer is \( \sqrt{3} \), the angle of elevation is \( 60^\circ \).

🎯 Exam Tip: When dealing with height and distance problems, always sketch a right triangle. Label the knowns and unknowns, then choose the appropriate trigonometric ratio (sine, cosine, or tangent) to solve for the required value.

Question 16. If \( \sin \theta = x \) and \( \sec \theta = y \), then find the value of \( \cot \theta \)
Answer: We are given the following:
1. \( \sin \theta = x \)
2. \( \sec \theta = y \)
We need to find the value of \( \cot \theta \).
From the definition of secant, we know that \( \sec \theta = \frac{1}{\cos \theta} \).
So, from the second given equation:
\( y = \frac{1}{\cos \theta} \)
We can rearrange this to find \( \cos \theta \):
\( \cos \theta = \frac{1}{y} \)
Now, we know that \( \cot \theta \) is defined as \( \frac{\cos \theta}{\sin \theta} \).
Substitute the values of \( \sin \theta \) and \( \cos \theta \) that we found:
\( \cot \theta = \frac{\frac{1}{y}}{x} \)
To simplify this complex fraction, multiply the numerator by the reciprocal of the denominator:
\( \cot \theta = \frac{1}{y} \times \frac{1}{x} \)
\( \cot \theta = \frac{1}{xy} \)
So, the value of \( \cot \theta \) is \( \frac{1}{xy} \).
In simple words: We are given sine and secant values. First, we use the secant value to find cosine. Then, we simply divide the cosine by the sine value to get cotangent, which is \( \frac{1}{xy} \).

🎯 Exam Tip: When given different trigonometric ratios and asked to find another, convert them all into terms of sine and cosine first. This often makes the calculation straightforward.

Question 17. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see figure).
Answer: Let's model this scenario as a right-angled triangle. The rope is the hypotenuse, and the pole is the vertical side opposite the angle formed with the ground.
Let the height of the pole be \( AB \).
Length of the rope (hypotenuse \( AC \)) \( = 20 \) m.
Angle made by the rope with the ground (angle of elevation) \( = 30^\circ \).
We use the sine function, which connects the height (opposite side) and the rope length (hypotenuse):
\( \sin (\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}} \)
\( \sin 30^\circ = \frac{AB}{AC} \)
Substitute the known values:
\( \sin 30^\circ = \frac{AB}{20} \)
We know that \( \sin 30^\circ = \frac{1}{2} \).
\( \frac{1}{2} = \frac{AB}{20} \)
To find \( AB \), multiply both sides by 20:
\( AB = \frac{1}{2} \times 20 \)
\( AB = 10 \) m.
Therefore, the height of the pole is 10 meters.
In simple words: The rope, pole, and ground form a right triangle. Given the rope's length and the angle it makes with the ground, we use the sine rule to find the pole's height.

🎯 Exam Tip: Always sketch the scenario to identify the hypotenuse, opposite, and adjacent sides. This helps in choosing the correct trigonometric ratio to solve for the unknown quantity.

Question 18. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Answer: Let's represent this problem using a right-angled triangle. The height of the kite is the vertical side (opposite to the angle of inclination), and the string length is the hypotenuse.
Let \( OB \) be the length of the string.
Height of the kite (opposite side \( AB \)) \( = 60 \) m.
Angle of inclination with the ground \( = 60^\circ \).
We use the sine function, which relates the opposite side and the hypotenuse:
\( \sin (\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}} \)
\( \sin 60^\circ = \frac{AB}{OB} \)
Substitute the known values:
\( \sin 60^\circ = \frac{60}{OB} \)
We know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
\( \frac{\sqrt{3}}{2} = \frac{60}{OB} \)
To solve for \( OB \), cross-multiply:
\( OB \cdot \sqrt{3} = 60 \cdot 2 \)
\( OB \cdot \sqrt{3} = 120 \)
\( OB = \frac{120}{\sqrt{3}} \)
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} \):
\( OB = \frac{120\sqrt{3}}{3} \)
\( OB = 40\sqrt{3} \) m.
Thus, the length of the string is \( 40\sqrt{3} \) meters.
In simple words: We have the kite's height and the angle of the string, forming a right triangle. Using the sine rule, we calculate the length of the string by dividing the height by \( \sin 60^\circ \).

🎯 Exam Tip: Remember to rationalize the denominator if your answer contains a square root in the bottom. This presents the answer in a standard mathematical form.

Question 19. Prove that \( \sin^6 \theta + \cos^6 \theta + 3\sin^2 \theta \cos^2 \theta = 1 \)
Answer: Let's simplify the left-hand side of the identity.
L.H.S \( = \sin^6 \theta + \cos^6 \theta + 3\sin^2 \theta \cos^2 \theta \)
We can rewrite \( \sin^6 \theta \) as \( (\sin^2 \theta)^3 \) and \( \cos^6 \theta \) as \( (\cos^2 \theta)^3 \).
So, the first two terms are in the form \( a^3 + b^3 \), where \( a = \sin^2 \theta \) and \( b = \cos^2 \theta \).
Recall the algebraic identity: \( a^3 + b^3 = (a+b)^3 – 3ab(a+b) \).
Substitute \( a = \sin^2 \theta \) and \( b = \cos^2 \theta \):
\( \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)^3 – 3(\sin^2 \theta)(\cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) \)
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = (1)^3 – 3\sin^2 \theta \cos^2 \theta (1) \)
\( = 1 – 3\sin^2 \theta \cos^2 \theta \)
Now, substitute this result back into the original L.H.S:
L.H.S \( = (1 – 3\sin^2 \theta \cos^2 \theta) + 3\sin^2 \theta \cos^2 \theta \)
The \( -3\sin^2 \theta \cos^2 \theta \) and \( +3\sin^2 \theta \cos^2 \theta \) terms cancel each other out:
\( = 1 \)
This matches the right-hand side of the identity, proving it.
In simple words: We treat \( \sin^6 \theta + \cos^6 \theta \) like \( a^3 + b^3 \). Using the formula for \( a^3 + b^3 \) and the rule \( \sin^2 \theta + \cos^2 \theta = 1 \), the expression simplifies greatly. All the extra terms cancel out, leaving just 1.

🎯 Exam Tip: For higher powers of trigonometric functions, try to express them as powers of \( \sin^2 \theta \) and \( \cos^2 \theta \), and then use algebraic identities like \( a^3+b^3 \) or \( a^2+b^2 \) along with \( \sin^2 \theta + \cos^2 \theta = 1 \).

Question 1. Prove that \( \sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}=2 \csc \theta \)
Answer: Let's simplify the left-hand side of the identity.
L.H.S \( = \sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}} \)
To combine these square root terms, find a common denominator under the radical signs by multiplying the numerator and denominator of each fraction by its conjugate. Alternatively, we can combine the two fractions into a single fraction:
\( = \frac{\sqrt{(\sec \theta-1)(\sec \theta-1)} + \sqrt{(\sec \theta+1)(\sec \theta+1)}}{\sqrt{(\sec \theta+1)(\sec \theta-1)}} \)
This simplifies to:
\( = \frac{(\sec \theta-1) + (\sec \theta+1)}{\sqrt{\sec^2 \theta-1^2}} \)
In the numerator, \( -1 \) and \( +1 \) cancel out, and the denominator simplifies using \( \sec^2 \theta - 1 = \tan^2 \theta \):
\( = \frac{2\sec \theta}{\sqrt{\tan^2 \theta}} \)
Assuming \( \theta \) is in a quadrant where \( \tan \theta \) is positive, so \( \sqrt{\tan^2 \theta} = \tan \theta \):
\( = \frac{2\sec \theta}{\tan \theta} \)
Now, convert \( \sec \theta \) and \( \tan \theta \) to sine and cosine:
\( = \frac{2 (\frac{1}{\cos \theta})}{\frac{\sin \theta}{\cos \theta}} \)
Multiply the numerator by the reciprocal of the denominator:
\( = \frac{2}{\cos \theta} \times \frac{\cos \theta}{\sin \theta} \)
The \( \cos \theta \) terms cancel out:
\( = \frac{2}{\sin \theta} \)
Using the reciprocal identity \( \frac{1}{\sin \theta} = \csc \theta \):
\( = 2\csc \theta \)
This matches the right-hand side of the identity, proving it.
In simple words: We combine the two square root terms into one big fraction. Then, we simplify the top and bottom. Using the rule \( \sec^2 \theta - 1 = \tan^2 \theta \) and changing to sine and cosine makes everything cancel out, leaving \( 2 \csc \theta \).

🎯 Exam Tip: When dealing with square roots of fractions, look for opportunities to use \( \sec^2 \theta - 1 = \tan^2 \theta \) or \( \csc^2 \theta - 1 = \cot^2 \theta \) as these transformations simplify the square root significantly.

Question 2. Prove that \( \frac{1+\cos A}{\sin A}+\frac{\sin A}{1+\cos A} = 2 \csc A \)
Answer: Let's simplify the left-hand side of the identity.
L.H.S \( = \frac{1+\cos A}{\sin A}+\frac{\sin A}{1+\cos A} \)
To add these fractions, find a common denominator, which is \( \sin A (1+\cos A) \):
\( = \frac{(1+\cos A)(1+\cos A) + \sin A \cdot \sin A}{\sin A (1+\cos A)} \)
\( = \frac{(1+\cos A)^2 + \sin^2 A}{\sin A (1+\cos A)} \)
Expand \( (1+\cos A)^2 \) as \( 1 + 2\cos A + \cos^2 A \):
\( = \frac{1 + 2\cos A + \cos^2 A + \sin^2 A}{\sin A (1+\cos A)} \)
Use the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \):
\( = \frac{1 + 2\cos A + 1}{\sin A (1+\cos A)} \)
Combine the constant terms:
\( = \frac{2 + 2\cos A}{\sin A (1+\cos A)} \)
Factor out 2 from the numerator:
\( = \frac{2(1 + \cos A)}{\sin A (1+\cos A)} \)
Cancel the common term \( (1 + \cos A) \) from the numerator and denominator:
\( = \frac{2}{\sin A} \)
Since \( \frac{1}{\sin A} = \csc A \), we have:
\( = 2\csc A \)
This matches the right-hand side of the identity, thus proving it.
In simple words: We combine the two fractions by finding a common bottom part. Expanding the top and using the rule \( \sin^2 A + \cos^2 A = 1 \) simplifies it to \( 2(1+\cos A) \) over \( \sin A (1+\cos A) \). Canceling the common term leaves \( 2/\sin A \), which is \( 2 \csc A \).

🎯 Exam Tip: When adding fractions, correctly finding the least common multiple of the denominators is essential. Then, remember to use Pythagorean identities to simplify the numerator after expansion.

Question 3. Prove that \( 2 (\sin^6 \theta + \cos^6 \theta) – 3 (\sin^4 \theta + \cos^4 \theta) + 1 = 0 \)
Answer: Let's simplify the left-hand side of the identity.
L.H.S \( = 2 (\sin^6 \theta + \cos^6 \theta) – 3 (\sin^4 \theta + \cos^4 \theta) + 1 \)

First, simplify \( \sin^6 \theta + \cos^6 \theta \):
Let \( a = \sin^2 \theta \) and \( b = \cos^2 \theta \). Then \( a^3+b^3 = (\sin^2 \theta)^3 + (\cos^2 \theta)^3 \).
Using the identity \( a^3 + b^3 = (a+b)^3 – 3ab(a+b) \):
\( \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)^3 – 3\sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta) \)
Using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = (1)^3 – 3\sin^2 \theta \cos^2 \theta (1) \)
\( = 1 – 3\sin^2 \theta \cos^2 \theta \)

Next, simplify \( \sin^4 \theta + \cos^4 \theta \):
Let \( a = \sin^2 \theta \) and \( b = \cos^2 \theta \). Then \( a^2+b^2 = (\sin^2 \theta)^2 + (\cos^2 \theta)^2 \).
Using the identity \( a^2 + b^2 = (a+b)^2 – 2ab \):
\( \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 – 2\sin^2 \theta \cos^2 \theta \)
Using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = (1)^2 – 2\sin^2 \theta \cos^2 \theta \)
\( = 1 – 2\sin^2 \theta \cos^2 \theta \)

Now substitute these simplified expressions back into the original L.H.S:
L.H.S \( = 2 (1 – 3\sin^2 \theta \cos^2 \theta) – 3 (1 – 2\sin^2 \theta \cos^2 \theta) + 1 \)
Distribute the 2 and -3:
\( = 2 – 6\sin^2 \theta \cos^2 \theta – 3 + 6\sin^2 \theta \cos^2 \theta + 1 \)
Group the constant terms and the \( \sin^2 \theta \cos^2 \theta \) terms:
\( = (2 – 3 + 1) + (-6\sin^2 \theta \cos^2 \theta + 6\sin^2 \theta \cos^2 \theta) \)
\( = 0 + 0 \)
\( = 0 \)
This matches the right-hand side of the identity, proving it.
In simple words: We first simplify \( \sin^6 \theta + \cos^6 \theta \) and \( \sin^4 \theta + \cos^4 \theta \) using algebraic formulas and the \( \sin^2 \theta + \cos^2 \theta = 1 \) rule. Then, we put these simpler forms back into the main equation. All the terms cancel out perfectly, leaving zero.

🎯 Exam Tip: When dealing with higher powers, remember to express them as powers of \( \sin^2 \theta \) and \( \cos^2 \theta \). This allows you to apply algebraic identities like \( (a+b)^3 \) or \( (a+b)^2 \) which, combined with \( \sin^2 \theta + \cos^2 \theta = 1 \), often simplify the expression significantly.

Question 4. Prove that \( \frac{\sin (90^\circ-\theta)}{1+\sin \theta}+\frac{\cos \theta}{1-\cos (90^\circ-\theta)} = 2 \sec \theta \)
Answer: Let's simplify the left-hand side of the identity.
L.H.S \( = \frac{\sin (90^\circ-\theta)}{1+\sin \theta}+\frac{\cos \theta}{1-\cos (90^\circ-\theta)} \)
Using the complementary angle identities, \( \sin (90^\circ – \theta) = \cos \theta \) and \( \cos (90^\circ – \theta) = \sin \theta \):
\( = \frac{\cos \theta}{1+\sin \theta} + \frac{\cos \theta}{1-\sin \theta} \)
To add these fractions, find a common denominator, which is \( (1+\sin \theta)(1-\sin \theta) \):
\( = \frac{\cos \theta (1-\sin \theta) + \cos \theta (1+\sin \theta)}{(1+\sin \theta)(1-\sin \theta)} \)
Expand the numerator:
\( = \frac{\cos \theta – \cos \theta \sin \theta + \cos \theta + \cos \theta \sin \theta}{1 – \sin^2 \theta} \)
In the numerator, \( -\cos \theta \sin \theta \) and \( +\cos \theta \sin \theta \) cancel out. In the denominator, use the Pythagorean identity \( 1 – \sin^2 \theta = \cos^2 \theta \):
\( = \frac{2\cos \theta}{\cos^2 \theta} \)
Cancel one \( \cos \theta \) term from the numerator and denominator:
\( = \frac{2}{\cos \theta} \)
Using the reciprocal identity \( \frac{1}{\cos \theta} = \sec \theta \):
\( = 2\sec \theta \)
This matches the right-hand side of the identity, thus proving it.
In simple words: First, we change the sine and cosine terms with \( 90^\circ-\theta \) to \( \cos \theta \) and \( \sin \theta \). Then, we add the fractions by finding a common bottom part. After simplifying the top and using the rule that \( 1 - \sin^2 \theta \) is \( \cos^2 \theta \), most terms cancel out, leaving \( 2 \sec \theta \).

🎯 Exam Tip: Always apply complementary angle identities like \( \sin(90^\circ-\theta) = \cos \theta \) as a first step to simplify expressions before combining fractions or applying other identities.

Question 5. Prove that \( \frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta} = 1 + \sec \theta \csc \theta \)
Answer: Let's simplify the left-hand side of the identity.
L.H.S \( = \frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta} \)
Convert all terms to sine and cosine:
\( = \frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}} \)
Simplify the denominators by finding a common denominator for them:
\( = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}} \)
Now, invert and multiply each fraction:
\( = \frac{\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\sin \theta-\cos \theta} + \frac{\cos \theta}{\sin \theta} \times \frac{\cos \theta}{\cos \theta-\sin \theta} \)
\( = \frac{\sin^2 \theta}{\cos \theta(\sin \theta-\cos \theta)} + \frac{\cos^2 \theta}{\sin \theta(-(\sin \theta-\cos \theta))} \)
Rewrite the second term to have a common denominator \( (\sin \theta-\cos \theta) \):
\( = \frac{\sin^2 \theta}{\cos \theta(\sin \theta-\cos \theta)} – \frac{\cos^2 \theta}{\sin \theta(\sin \theta-\cos \theta)} \)
Combine the fractions with the common denominator:
\( = \frac{1}{\sin \theta-\cos \theta} \left( \frac{\sin^2 \theta}{\cos \theta} – \frac{\cos^2 \theta}{\sin \theta} \right) \)
Combine the terms inside the parenthesis:
\( = \frac{1}{\sin \theta-\cos \theta} \left( \frac{\sin^3 \theta-\cos^3 \theta}{\sin \theta \cos \theta} \right) \)
Use the algebraic identity \( a^3-b^3 = (a-b)(a^2+ab+b^2) \), where \( a=\sin \theta \) and \( b=\cos \theta \):
\( = \frac{1}{\sin \theta-\cos \theta} \left( \frac{(\sin \theta-\cos \theta)(\sin^2 \theta+\sin \theta \cos \theta+\cos^2 \theta)}{\sin \theta \cos \theta} \right) \)
Cancel out \( (\sin \theta-\cos \theta) \) and use \( \sin^2 \theta+\cos^2 \theta = 1 \):
\( = \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} \)
Separate the fraction into two terms:
\( = \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta} \)
\( = \frac{1}{\sin \theta} \cdot \frac{1}{\cos \theta} + 1 \)
Using reciprocal identities, \( \frac{1}{\sin \theta} = \csc \theta \) and \( \frac{1}{\cos \theta} = \sec \theta \):
\( = \csc \theta \sec \theta + 1 \)
\( = 1 + \sec \theta \csc \theta \)
This matches the right-hand side of the identity, proving it.
In simple words: This problem involves lots of steps! We change everything to sine and cosine. Then, we simplify the fractions and combine them. Using the \( a^3-b^3 \) formula and the \( \sin^2 \theta + \cos^2 \theta = 1 \) rule helps us cancel out big parts, leading to \( 1 + \sec \theta \csc \theta \).

🎯 Exam Tip: For complex identities involving tangent and cotangent, converting everything to sine and cosine is a robust strategy. Be careful with algebraic identities like \( a^3-b^3 \) and managing denominators.

Question 6. Prove that \( \frac{\sin (90^\circ-\theta)}{1-\tan \theta}+\frac{\cos (90^\circ-\theta)}{1-\cot \theta} = \cos \theta + \sin \theta \)
Answer: Let's simplify the left-hand side of the identity.
L.H.S \( = \frac{\sin (90^\circ-\theta)}{1-\tan \theta}+\frac{\cos (90^\circ-\theta)}{1-\cot \theta} \)
Using the complementary angle identities, \( \sin (90^\circ – \theta) = \cos \theta \) and \( \cos (90^\circ – \theta) = \sin \theta \):
\( = \frac{\cos \theta}{1-\tan \theta} + \frac{\sin \theta}{1-\cot \theta} \)
Now, convert \( \tan \theta \) and \( \cot \theta \) to sine and cosine:
\( = \frac{\cos \theta}{1-\frac{\sin \theta}{\cos \theta}} + \frac{\sin \theta}{1-\frac{\cos \theta}{\sin \theta}} \)
Simplify the denominators:
\( = \frac{\cos \theta}{\frac{\cos \theta-\sin \theta}{\cos \theta}} + \frac{\sin \theta}{\frac{\sin \theta-\cos \theta}{\sin \theta}} \)
Invert and multiply each term:
\( = \frac{\cos^2 \theta}{\cos \theta-\sin \theta} + \frac{\sin^2 \theta}{\sin \theta-\cos \theta} \)
To get a common denominator, rewrite \( (\sin \theta-\cos \theta) \) as \( -(\cos \theta-\sin \theta) \):
\( = \frac{\cos^2 \theta}{\cos \theta-\sin \theta} – \frac{\sin^2 \theta}{\cos \theta-\sin \theta} \)
Combine the fractions:
\( = \frac{\cos^2 \theta-\sin^2 \theta}{\cos \theta-\sin \theta} \)
Use the algebraic identity \( a^2-b^2 = (a-b)(a+b) \) in the numerator:
\( = \frac{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}{\cos \theta-\sin \theta} \)
Cancel out the common term \( (\cos \theta-\sin \theta) \):
\( = \cos \theta+\sin \theta \)
This matches the right-hand side of the identity, proving it.
In simple words: First, we change the angles to \( \theta \). Then we change tangent and cotangent to sine and cosine. After combining fractions and simplifying, we use the \( a^2-b^2 \) rule. This helps to cancel out parts of the fraction, leaving just \( \cos \theta + \sin \theta \).

🎯 Exam Tip: When terms like \( (a-b) \) and \( (b-a) \) appear in denominators, factor out \( -1 \) from one of them to make the denominators identical, which simplifies combining the fractions.

Question 7. Prove that \( \frac{\tan (90^\circ-\theta)}{\csc \theta+1}+\frac{\csc \theta+1}{\cot \theta} = 2 \sec \theta \)
Answer: Let's simplify the left-hand side of the identity.
L.H.S \( = \frac{\tan (90^\circ-\theta)}{\csc \theta+1}+\frac{\csc \theta+1}{\cot \theta} \)
Using the complementary angle identity \( \tan (90^\circ – \theta) = \cot \theta \):
\( = \frac{\cot \theta}{\csc \theta+1}+\frac{\csc \theta+1}{\cot \theta} \)
To combine these fractions, find a common denominator, which is \( (\csc \theta+1)\cot \theta \):
\( = \frac{\cot \theta \cdot \cot \theta + (\csc \theta+1)(\csc \theta+1)}{(\csc \theta+1)\cot \theta} \)
\( = \frac{\cot^2 \theta + (\csc \theta+1)^2}{(\csc \theta+1)\cot \theta} \)
Now, let's follow the solution's approach of simplifying the first term before combining, which is more straightforward in this case. The solution implicitly does this by multiplying the first fraction by its conjugate:
\( \frac{\cot \theta}{\csc \theta+1} = \frac{\cot \theta(\csc \theta-1)}{(\csc \theta+1)(\csc \theta-1)} = \frac{\cot \theta(\csc \theta-1)}{\csc^2 \theta-1} \)
Using the identity \( \csc^2 \theta-1 = \cot^2 \theta \):
\( = \frac{\cot \theta(\csc \theta-1)}{\cot^2 \theta} \)
Cancel one \( \cot \theta \) term from numerator and denominator:
\( = \frac{\csc \theta-1}{\cot \theta} \)
Now, substitute this simplified term back into the original L.H.S:
L.H.S \( = \frac{\csc \theta-1}{\cot \theta}+\frac{\csc \theta+1}{\cot \theta} \)
Since they now have a common denominator, combine the numerators:
\( = \frac{(\csc \theta-1) + (\csc \theta+1)}{\cot \theta} \)
In the numerator, \( -1 \) and \( +1 \) cancel out:
\( = \frac{2\csc \theta}{\cot \theta} \)
Now, convert \( \csc \theta \) and \( \cot \theta \) to sine and cosine:
\( = \frac{2 (\frac{1}{\sin \theta})}{\frac{\cos \theta}{\sin \theta}} \)
Multiply the numerator by the reciprocal of the denominator:
\( = \frac{2}{\sin \theta} \times \frac{\sin \theta}{\cos \theta} \)
The \( \sin \theta \) terms cancel out:
\( = \frac{2}{\cos \theta} \)
Using the reciprocal identity \( \frac{1}{\cos \theta} = \sec \theta \):
\( = 2\sec \theta \)
This matches the right-hand side of the identity, proving it.
In simple words: First, we change \( \tan (90^\circ-\theta) \) to \( \cot \theta \). Then, we simplify the first fraction by multiplying its top and bottom by \( \csc \theta-1 \). This allows us to use the rule \( \csc^2 \theta-1 = \cot^2 \theta \). After simplification, both fractions have the same bottom part, so we combine them. Finally, we convert to sine and cosine, and everything cancels out to \( 2 \sec \theta \).

🎯 Exam Tip: When dealing with sum/difference of fractions, sometimes simplifying each fraction individually before combining them can be more efficient than finding a single common denominator immediately for the entire expression.

Question 8. Prove that \( \frac{\cot \theta + \operatorname{cosec} \theta-1}{\cot \theta - \operatorname{cosec} \theta + 1} = \operatorname{cosec} \theta + \cot \theta \)
Answer:
L.H.S \( = \frac{\cot \theta + \operatorname{cosec} \theta-1}{\cot \theta - \operatorname{cosec} \theta + 1} \)
We know the identity: \( \operatorname{cosec}^2 \theta - \cot^2 \theta = 1 \). We will replace the '1' in the numerator with this identity.
\( = \frac{(\cot \theta + \operatorname{cosec} \theta) - (\operatorname{cosec}^2 \theta - \cot^2 \theta)}{\cot \theta - \operatorname{cosec} \theta + 1} \)
Now, factor the term \( \operatorname{cosec}^2 \theta - \cot^2 \theta \) using \( (a^2 - b^2) = (a-b)(a+b) \):
\( = \frac{(\cot \theta + \operatorname{cosec} \theta) - [(\operatorname{cosec} \theta - \cot \theta)(\operatorname{cosec} \theta + \cot \theta)]}{\cot \theta - \operatorname{cosec} \theta + 1} \)
Factor out the common term \( (\cot \theta + \operatorname{cosec} \theta) \) from the numerator:
\( = \frac{(\cot \theta + \operatorname{cosec} \theta) [1 - (\operatorname{cosec} \theta - \cot \theta)]}{\cot \theta - \operatorname{cosec} \theta + 1} \)
Simplify the expression inside the square brackets:
\( = \frac{(\cot \theta + \operatorname{cosec} \theta) [1 - \operatorname{cosec} \theta + \cot \theta]}{\cot \theta - \operatorname{cosec} \theta + 1} \)
Observe that the term \( (1 - \operatorname{cosec} \theta + \cot \theta) \) in the numerator is exactly the same as the denominator \( (\cot \theta - \operatorname{cosec} \theta + 1) \). So, these terms cancel each other out.
\( = \cot \theta + \operatorname{cosec} \theta \)
Thus, L.H.S = R.H.S.
In simple words: To prove this identity, we start with the left side. We use the identity \( \operatorname{cosec}^2 \theta - \cot^2 \theta = 1 \) to rewrite the number 1 in the numerator. Then, we factor out a common term from the top part. After simplifying, the remaining expression matches the right side, showing the identity is true.

🎯 Exam Tip: When proving identities, look for ways to use fundamental identities like \( \operatorname{cosec}^2 \theta - \cot^2 \theta = 1 \) or factor common terms to simplify expressions, especially when there's a constant term (like 1) that can be replaced by an identity.

Question 9. Prove that \( (1 + \cot \theta – \operatorname{cosec} \theta) (1 + \tan \theta + \sec \theta) = 2 \)
Answer:
L.H.S \( = (1 + \cot \theta – \operatorname{cosec} \theta) (1 + \tan \theta + \sec \theta) \)
To simplify, convert all trigonometric ratios into \( \sin \theta \) and \( \cos \theta \):
\( \cot \theta = \frac{\cos \theta}{\sin \theta} \), \( \operatorname{cosec} \theta = \frac{1}{\sin \theta} \), \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), \( \sec \theta = \frac{1}{\cos \theta} \)
Substitute these into the expression:
\( = \left(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right) \left(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right) \)
Combine the terms within each parenthesis by finding a common denominator:
\( = \left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right) \left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right) \)
Rearrange the terms in the numerator slightly to see a pattern. Let \( A = (\sin \theta + \cos \theta) \) and \( B = 1 \). The numerators are then \( (A - B) \) and \( (A + B) \).
\( = \frac{(\sin \theta + \cos \theta - 1)(\sin \theta + \cos \theta + 1)}{\sin \theta \cos \theta} \)
Use the algebraic identity \( (a-b)(a+b) = a^2 - b^2 \) for the numerator:
\( = \frac{(\sin \theta + \cos \theta)^2 - 1^2}{\sin \theta \cos \theta} \)
Expand \( (\sin \theta + \cos \theta)^2 \):
\( = \frac{\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta - 1}{\sin \theta \cos \theta} \)
Apply the fundamental identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = \frac{1 + 2 \sin \theta \cos \theta - 1}{\sin \theta \cos \theta} \)
Simplify the numerator:
\( = \frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta} \)
Cancel out the common term \( \sin \theta \cos \theta \):
\( = 2 \)
Thus, L.H.S = R.H.S.
In simple words: To prove this, we change all cot, cosec, tan, and sec terms into sin and cos. We then combine the fractions by finding a common bottom part. Using the formula \( (a-b)(a+b) = a^2 - b^2 \) and the basic identity \( \sin^2 \theta + \cos^2 \theta = 1 \), the expression simplifies down to 2, which is the right side.

🎯 Exam Tip: When an identity involves many different trigonometric functions, converting all terms to \( \sin \) and \( \cos \) is often the first and most effective step. Look for opportunities to use algebraic identities like \( (a+b)^2 \) or \( (a-b)(a+b) \).

Question 10. Prove that \( \frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta} \)
Answer:
L.H.S \( = \frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1} \)
To introduce \( \tan \theta \) and \( \sec \theta \), divide every term in the numerator and denominator by \( \cos \theta \):
\( = \frac{\frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\cos \theta} + \frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\cos \theta} - \frac{1}{\cos \theta}} \)
\( = \frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta} \)
Rearrange the terms in the numerator to group \( \sec \theta \) and \( \tan \theta \):
\( = \frac{(\sec \theta + \tan \theta) - 1}{(\tan \theta - \sec \theta) + 1} \)
Replace \( 1 \) in the numerator with the identity \( \sec^2 \theta - \tan^2 \theta \) (since \( \sec^2 \theta - \tan^2 \theta = 1 \)):
\( = \frac{(\sec \theta + \tan \theta) - (\sec^2 \theta - \tan^2 \theta)}{\tan \theta - \sec \theta + 1} \)
Factor \( (\sec^2 \theta - \tan^2 \theta) \) using the algebraic identity \( (a^2 - b^2) = (a-b)(a+b) \):
\( = \frac{(\sec \theta + \tan \theta) - [(\sec \theta - \tan \theta)(\sec \theta + \tan \theta)]}{\tan \theta - \sec \theta + 1} \)
Factor out the common term \( (\sec \theta + \tan \theta) \) from the numerator:
\( = \frac{(\sec \theta + \tan \theta) [1 - (\sec \theta - \tan \theta)]}{\tan \theta - \sec \theta + 1} \)
Simplify the term inside the square brackets:
\( = \frac{(\sec \theta + \tan \theta) [1 - \sec \theta + \tan \theta]}{\tan \theta - \sec \theta + 1} \)
Notice that \( (1 - \sec \theta + \tan \theta) \) is the same as the denominator \( (\tan \theta - \sec \theta + 1) \). So, these terms cancel out.
\( = \sec \theta + \tan \theta \)
Now, recall the identity \( \sec^2 \theta - \tan^2 \theta = 1 \). We can write this as:
\( (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1 \)
Dividing both sides by \( (\sec \theta - \tan \theta) \), we get:
\( \sec \theta + \tan \theta = \frac{1}{\sec \theta - \tan \theta} \)
Therefore, L.H.S = R.H.S.
In simple words: To prove this, we first divide everything by \( \cos \theta \) to get \( \tan \theta \) and \( \sec \theta \) terms. Then, we use the identity \( \sec^2 \theta - \tan^2 \theta = 1 \) to rewrite the number 1 in the numerator. This helps us factor the top part. After cancelling similar terms, we are left with \( \sec \theta + \tan \theta \), which can be further changed to \( \frac{1}{\sec \theta - \tan \theta} \) using the same identity.

🎯 Exam Tip: When a question involves \( \sin \theta \) and \( \cos \theta \) along with \( +1 \) or \( -1 \) in the numerator or denominator, dividing by \( \cos \theta \) (or \( \sin \theta \) if appropriate) and then using \( \sec^2 \theta - \tan^2 \theta = 1 \) or \( \operatorname{cosec}^2 \theta - \cot^2 \theta = 1 \) is often the key. This technique is also known as rationalization by identity.

Question 16. If \( \sin \theta = x \) and \( \sec \theta = y \), then find the value of \( \cot \theta \)
Answer:
We are given:
\( \sin \theta = x \)
\( \sec \theta = y \)
We know that \( \sec \theta \) is the reciprocal of \( \cos \theta \). So, \( \sec \theta = \frac{1}{\cos \theta} \).
Substituting the given value:
\( y = \frac{1}{\cos \theta} \)
Rearranging this, we find \( \cos \theta \):
\( \cos \theta = \frac{1}{y} \)
We need to find the value of \( \cot \theta \). The definition of \( \cot \theta \) is \( \frac{\cos \theta}{\sin \theta} \).
Substitute the values we have for \( \sin \theta \) and \( \cos \theta \):
\( \cot \theta = \frac{\frac{1}{y}}{x} \)
To simplify the complex fraction, multiply the denominator \( x \) by \( y \):
\( \cot \theta = \frac{1}{xy} \)
The value of \( \cot \theta \) in terms of \( x \) and \( y \) is \( \frac{1}{xy} \).
In simple words: We are given \( \sin \theta \) and \( \sec \theta \). We need to find \( \cot \theta \). First, we find \( \cos \theta \) from \( \sec \theta \). Then, we use the formula that \( \cot \theta \) is \( \cos \theta \) divided by \( \sin \theta \). Putting the given values in gives us the answer.

🎯 Exam Tip: When asked to find one trigonometric ratio from others, convert all given terms into \( \sin \) and \( \cos \) as much as possible. This makes it easier to use basic identities to find the required ratio.

Question 17. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° ( see figure).
Answer:
Let AB represent the vertical pole, and AC represent the rope.
The length of the rope, \( AC = 20 \text{ m} \).
The angle the rope makes with the ground is \( \angle ACB = 30^\circ \).
In the right-angled triangle ABC (right-angled at B, at the base of the pole):
We need to find the height of the pole, which is AB.
We can use the sine function, as it relates the side opposite to the angle (AB) to the hypotenuse (AC).
The formula for sine is: \( \sin(\text{angle}) = \frac{\text{Opposite}}{\text{Hypotenuse}} \)
So, \( \sin 30^\circ = \frac{AB}{AC} \)
We know that the value of \( \sin 30^\circ \) is \( \frac{1}{2} \).
Substitute the known values into the equation:
\( \frac{1}{2} = \frac{AB}{20} \)
To solve for AB, multiply both sides by 20:
\( AB = 20 \times \frac{1}{2} \)
\( AB = 10 \text{ m} \)
Therefore, the height of the pole is 10 meters.
In simple words: We have a right triangle formed by the pole, the ground, and the rope. The rope is 20 meters long and makes a 30-degree angle with the ground. We use the sine function, which connects the height of the pole (opposite side) to the rope's length (hypotenuse). Since \( \sin 30^\circ \) is 1/2, the pole's height is half the rope's length.

🎯 Exam Tip: Always draw a clear diagram for height and distance problems. Label the knowns and unknowns, then choose the correct trigonometric ratio (sine, cosine, or tangent) based on the sides involved in the problem.

Question 18. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Answer:
Let AB be the height of the kite from the ground. So, \( AB = 60 \text{ m} \).
Let OB be the length of the string.
The string is tied to a point O on the ground. The angle of inclination of the string with the ground is \( \angle AOB = 60^\circ \).
In the right-angled triangle OAB (right-angled at A):
We need to find the length of the string, OB.
We use the sine function, as it relates the opposite side (AB, the height) to the hypotenuse (OB, the string length).
The formula is: \( \sin(\text{angle}) = \frac{\text{Opposite}}{\text{Hypotenuse}} \)
So, \( \sin 60^\circ = \frac{AB}{OB} \)
We know that the value of \( \sin 60^\circ \) is \( \frac{\sqrt{3}}{2} \).
Substitute the known values:
\( \frac{\sqrt{3}}{2} = \frac{60}{OB} \)
To solve for OB, cross-multiply:
\( \sqrt{3} \times OB = 60 \times 2 \)
\( OB = \frac{120}{\sqrt{3}} \)
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} \):
\( OB = \frac{120 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \)
\( OB = \frac{120 \sqrt{3}}{3} \)
\( OB = 40 \sqrt{3} \text{ m} \)
Thus, the length of the string is \( 40 \sqrt{3} \) meters.
In simple words: The kite is 60 meters high, and its string makes a 60-degree angle with the ground. We need to find the length of the string. Using the sine function, which connects height to string length, we can set up an equation. Solving this equation gives us the string's length as \( 40 \sqrt{3} \) meters.

🎯 Exam Tip: Remember to rationalize the denominator if your answer contains a square root in the bottom. This makes the answer cleaner and easier to compare.

Question 19. Prove that \( \sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta = 1 \)
Answer:
L.H.S \( = \sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta \)
Rewrite \( \sin^6 \theta \) as \( (\sin^2 \theta)^3 \) and \( \cos^6 \theta \) as \( (\cos^2 \theta)^3 \):
\( = (\sin^2 \theta)^3 + (\cos^2 \theta)^3 + 3 \sin^2 \theta \cos^2 \theta \)
Use the algebraic identity for the sum of cubes: \( a^3 + b^3 = (a+b)^3 - 3ab(a+b) \).
Here, let \( a = \sin^2 \theta \) and \( b = \cos^2 \theta \).
\( = [(\sin^2 \theta + \cos^2 \theta)^3 - 3 \sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta)] + 3 \sin^2 \theta \cos^2 \theta \)
Apply the fundamental trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = [(1)^3 - 3 \sin^2 \theta \cos^2 \theta (1)] + 3 \sin^2 \theta \cos^2 \theta \)
Simplify the expression:
\( = [1 - 3 \sin^2 \theta \cos^2 \theta] + 3 \sin^2 \theta \cos^2 \theta \)
The terms \( -3 \sin^2 \theta \cos^2 \theta \) and \( +3 \sin^2 \theta \cos^2 \theta \) cancel each other out:
\( = 1 \)
Thus, L.H.S = R.H.S.
In simple words: To prove this identity, we first rewrite \( \sin^6 \theta \) and \( \cos^6 \theta \) as cubes of \( \sin^2 \theta \) and \( \cos^2 \theta \). Then we use a special algebra formula for \( a^3 + b^3 \). Since \( \sin^2 \theta + \cos^2 \theta \) is always 1, many terms cancel out, leaving us with just 1.

🎯 Exam Tip: For higher powers of \( \sin \) and \( \cos \), often express them in terms of squares and then use algebraic identities like \( (a+b)^3 \) or \( (a^3+b^3) \). Remember the fundamental identity \( \sin^2 \theta + \cos^2 \theta = 1 \) and how it simplifies expressions.

Question 13. If \( \sin \theta, \cos \theta \) and \( \tan \theta \) are in G. P., then prove that \( \cot^6 \theta – \cot^2 \theta = 1 \).
Answer:
Given that \( \sin \theta, \cos \theta, \tan \theta \) are in Geometric Progression (G.P.).
In a G.P., the square of the middle term is equal to the product of the first and last terms:
\( (\cos \theta)^2 = \sin \theta \times \tan \theta \)
\( \cos^2 \theta = \sin \theta \times \frac{\sin \theta}{\cos \theta} \)
\( \cos^2 \theta = \frac{\sin^2 \theta}{\cos \theta} \)
Multiply both sides by \( \cos \theta \):
\( \cos^3 \theta = \sin^2 \theta \) (Equation 1)
Now, we need to prove that \( \cot^6 \theta – \cot^2 \theta = 1 \).
L.H.S \( = \cot^6 \theta – \cot^2 \theta \)
Substitute \( \cot \theta = \frac{\cos \theta}{\sin \theta} \):
\( = \left(\frac{\cos \theta}{\sin \theta}\right)^6 - \left(\frac{\cos \theta}{\sin \theta}\right)^2 \)
\( = \frac{\cos^6 \theta}{\sin^6 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta} \)
Factor out \( \frac{\cos^2 \theta}{\sin^2 \theta} \):
\( = \frac{\cos^2 \theta}{\sin^2 \theta} \left(\frac{\cos^4 \theta}{\sin^4 \theta} - 1\right) \)
From Equation 1, we have \( \sin^2 \theta = \cos^3 \theta \). Substitute this into the expression:
\( = \frac{\cos^2 \theta}{\cos^3 \theta} \left(\frac{\cos^4 \theta}{(\cos^3 \theta)^2} - 1\right) \)
Simplify the powers:
\( = \frac{1}{\cos \theta} \left(\frac{\cos^4 \theta}{\cos^6 \theta} - 1\right) \)
\( = \frac{1}{\cos \theta} \left(\frac{1}{\cos^2 \theta} - 1\right) \)
Combine the terms inside the parenthesis:
\( = \frac{1}{\cos \theta} \left(\frac{1 - \cos^2 \theta}{\cos^2 \theta}\right) \)
Use the identity \( 1 - \cos^2 \theta = \sin^2 \theta \):
\( = \frac{1}{\cos \theta} \left(\frac{\sin^2 \theta}{\cos^2 \theta}\right) \)
Substitute \( \sin^2 \theta = \cos^3 \theta \) again:
\( = \frac{1}{\cos \theta} \left(\frac{\cos^3 \theta}{\cos^2 \theta}\right) \)
Simplify the expression:
\( = \frac{1}{\cos \theta} \times \cos \theta \)
\( = 1 \)
Thus, L.H.S = R.H.S.
In simple words: First, we use the property of Geometric Progression, which is \( b^2 = ac \). This gives us a key relationship: \( \cos^3 \theta = \sin^2 \theta \). Then we start with the left side of the equation we need to prove, which is \( \cot^6 \theta – \cot^2 \theta \). We change \( \cot \theta \) to \( \frac{\cos \theta}{\sin \theta} \) and use our derived relationship to simplify the expression step by step. Everything cancels out to 1, proving the identity.

🎯 Exam Tip: When terms are in G.P., immediately write down the relationship between them (\( \text{middle}^2 = \text{first} \times \text{last} \)). This derived relationship is crucial for simplifying the expression you need to prove in the second part of the question.

Question 14. A person in an helicopter flying at a height of 700 m, observes two objects lying opposite to each other on either bank of a river. The angles of depression of the objects are 30° and 45°. Find the width of the river. \( (\sqrt { 3 } = 1. 732) \)
Answer:
Let C be the position of the helicopter and CD be its height above the river. So, \( CD = 700 \text{ m} \).
Let A and B be the two objects on opposite banks of the river. D is the point directly below the helicopter on the river.
The angle of depression to object A is \( 30^\circ \). By alternate interior angles, \( \angle CAD = 30^\circ \).
The angle of depression to object B is \( 45^\circ \). By alternate interior angles, \( \angle CBD = 45^\circ \).
The width of the river is \( AB = AD + BD \).
First, consider the right-angled triangle ACD:
\( \tan 30^\circ = \frac{CD}{AD} \)
We know \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
\( \frac{1}{\sqrt{3}} = \frac{700}{AD} \)
Cross-multiplying gives:
\( AD = 700 \sqrt{3} \text{ m} \)
Next, consider the right-angled triangle BCD:
\( \tan 45^\circ = \frac{CD}{BD} \)
We know \( \tan 45^\circ = 1 \).
\( 1 = \frac{700}{BD} \)
This means:
\( BD = 700 \text{ m} \)
Now, calculate the total width of the river by adding AD and BD:
\( AB = AD + BD = 700 \sqrt{3} + 700 \)
Factor out 700:
\( AB = 700 (\sqrt{3} + 1) \)
Using the given value \( \sqrt{3} = 1.732 \):
\( AB = 700 (1.732 + 1) \)
\( AB = 700 (2.732) \)
\( AB = 1912.4 \text{ m} \)
The width of the river is approximately 1912.4 meters.
In simple words: A helicopter is 700 meters high. It sees two things on opposite sides of a river at angles of 30 and 45 degrees. We use the tangent function to find how far each thing is from the point directly below the helicopter. One side is \( 700 \sqrt{3} \) meters, and the other is 700 meters. Adding these together gives the river's total width.

🎯 Exam Tip: When angles of depression are given, remember that the angle of elevation from the object to the observer is the same. Draw two clear right triangles sharing a common height, and use the tangent function to find horizontal distances, then add them if the objects are on opposite sides.

Question 15. A person X standing on a horizontal plane, observes a bird flying at a distance of 100 m from him at an angle of elevation of 30°. Another person Y standing on the roof of a 20 m high building, observes the bird at the same time at an angle of elevation of 45°. If X and Y are on the opposite sides of the bird, then find the distance of the bird from Y.
Answer:
Let B be the position of the bird.
Let X be the first person on the ground. The distance from X to the bird is \( BX = 100 \text{ m} \).
The angle of elevation of the bird from X is \( 30^\circ \), so \( \angle BXD = 30^\circ \).
Let BD be the vertical height of the bird from the ground.
In the right-angled triangle BXD (right-angled at D):
\( \sin 30^\circ = \frac{BD}{BX} \)
\( \frac{1}{2} = \frac{BD}{100} \)
Solving for BD:
\( BD = \frac{100}{2} = 50 \text{ m} \)
So, the height of the bird from the ground is 50 m.
Let Y be the second person standing on the roof of a 20 m high building.
Let A be the base of the building, and C be the point on the vertical line directly below the bird, at the same height as Y's eye level (which is 20 m above ground).
The height of the building \( AY = 20 \text{ m} \).
The effective height of the bird from Y's eye level (BC) will be the bird's total height minus the building's height:
\( BC = BD - AY = 50 - 20 = 30 \text{ m} \).
The angle of elevation of the bird from Y is \( 45^\circ \), so \( \angle BYC = 45^\circ \).
In the right-angled triangle BCY (right-angled at C):
We need to find BY, which is the distance of the bird from person Y.
\( \sin 45^\circ = \frac{BC}{BY} \)
\( \frac{1}{\sqrt{2}} = \frac{30}{BY} \)
Solving for BY:
\( BY = 30 \sqrt{2} \text{ m} \)
The distance of the bird from person Y is \( 30 \sqrt{2} \) meters.
In simple words: Person X sees a bird 100 meters away at a 30-degree angle. This helps us find the bird's actual height, which is 50 meters. Person Y is on a 20-meter building and sees the same bird at a 45-degree angle. We calculate the bird's height from Y's eye level (50 - 20 = 30 meters). Then, using the sine function for person Y's view, we find the distance from Y to the bird.

🎯 Exam Tip: When there are multiple observers or different height levels, first find the common height (or distance). Then, carefully adjust the heights to the relevant eye level for each observation to set up the correct triangles and apply trigonometric ratios.

Question 16. A boy is standing at some distance from a 30 m tall building and his eye level from the ground is 1.5 m. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Answer:
Let P be the top of the building and Q be the base of the building.
The total height of the building is \( PQ = 30 \text{ m} \).
The boy's eye level is at 1.5 m from the ground. So, the effective height of the building from his eye level is \( PQ' = 30 - 1.5 = 28.5 \text{ m} \).
Let R be the initial position of the boy's eyes, and S be his final position after walking towards the building.
The initial angle of elevation from R to the top P is \( \angle PRQ' = 30^\circ \).
The final angle of elevation from S to the top P is \( \angle PSQ' = 60^\circ \).
Let the distance from the final position S to the base Q' be \( y \). So, \( SQ' = y \).
Let the distance the boy walked be \( x \). So, \( RS = x \).
The initial total distance from R to Q' is \( RQ' = RS + SQ' = x+y \).
Consider the right-angled triangle \( \triangle PQ'S \) (right-angled at Q'):
\( \tan 60^\circ = \frac{PQ'}{SQ'} \)
\( \sqrt{3} = \frac{28.5}{y} \)
From this, we find \( y \):
\( y = \frac{28.5}{\sqrt{3}} \) (Equation 1)
Now, consider the right-angled triangle \( \triangle PQ'R \) (right-angled at Q'):
\( \tan 30^\circ = \frac{PQ'}{RQ'} \)
\( \frac{1}{\sqrt{3}} = \frac{28.5}{x+y} \)
From this, we find \( x+y \):
\( x+y = 28.5 \sqrt{3} \) (Equation 2)
Substitute the value of \( y \) from Equation 1 into Equation 2:
\( x + \frac{28.5}{\sqrt{3}} = 28.5 \sqrt{3} \)
To find \( x \), rearrange the equation:
\( x = 28.5 \sqrt{3} - \frac{28.5}{\sqrt{3}} \)
Factor out 28.5:
\( x = 28.5 \left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) \)
Find a common denominator inside the parenthesis:
\( x = 28.5 \left(\frac{(\sqrt{3} \times \sqrt{3}) - 1}{\sqrt{3}}\right) \)
\( x = 28.5 \left(\frac{3 - 1}{\sqrt{3}}\right) \)
\( x = 28.5 \left(\frac{2}{\sqrt{3}}\right) \)
\( x = \frac{57}{\sqrt{3}} \)
Rationalize the denominator by multiplying by \( \frac{\sqrt{3}}{\sqrt{3}} \):
\( x = \frac{57 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \)
\( x = \frac{57 \sqrt{3}}{3} \)
\( x = 19 \sqrt{3} \text{ m} \)
The distance the boy walked towards the building is \( 19 \sqrt{3} \) meters.
In simple words: A boy watches a 30-meter building. His eyes are 1.5 meters from the ground, so he looks at the top 28.5 meters above his eye level. When he is far, the angle to the top is 30 degrees. As he walks closer, it becomes 60 degrees. We use the tangent function to set up two equations connecting distances and heights. By solving these equations, we find that he walked \( 19 \sqrt{3} \) meters.

🎯 Exam Tip: For problems involving a change in observation point or angle, draw two right triangles within one diagram. Calculate the effective height from the observer's eye level. Use the tangent function to relate heights and distances, then solve the system of equations formed.

Question 17. A straight highway leads to the foot of a tower. A man standing on the top of the tower spots a van at an angle of depression of 30°. The van is approaching the tower with a uniform speed. After 6 minutes, the angle of depression of the van is found to be 60°. How many more minutes will it take for the van to reach the tower?
Answer:
Let AD be the height of the tower, and let its height be \( h \).
Let A be the top of the tower, and D be its foot.
Let B be the initial position of the van, and C be its position after 6 minutes.
The angle of depression from A to B is \( 30^\circ \). By alternate interior angles, the angle of elevation from B to A is \( \angle ADB = 30^\circ \).
The angle of depression from A to C is \( 60^\circ \). By alternate interior angles, the angle of elevation from C to A is \( \angle ADC = 60^\circ \).
Let the uniform speed of the van be \( s \) meters per minute.
The distance the van traveled from B to C is \( BC = \text{speed} \times \text{time} = s \times 6 = 6s \) meters.
Let \( t \) be the time (in minutes) taken for the van to travel from C to D.
So, the distance \( CD = \text{speed} \times \text{time} = s \times t = st \) meters.
The total distance from the initial position B to the foot of the tower D is \( DB = DC + CB = st + 6s \).
Consider the right-angled triangle \( \triangle ADC \) (right-angled at D):
\( \tan 60^\circ = \frac{AD}{CD} \)
\( \sqrt{3} = \frac{h}{st} \)
From this, we get an expression for \( h \):
\( h = \sqrt{3} st \) (Equation 1)
Now, consider the right-angled triangle \( \triangle ADB \) (right-angled at D):
\( \tan 30^\circ = \frac{AD}{DB} \)
\( \frac{1}{\sqrt{3}} = \frac{h}{st + 6s} \)
From this, we get another expression for \( h \):
\( h = \frac{st + 6s}{\sqrt{3}} \) (Equation 2)
Equate the two expressions for \( h \) (from Equation 1 and Equation 2) to solve for \( t \):
\( \sqrt{3} st = \frac{st + 6s}{\sqrt{3}} \)
Multiply both sides by \( \sqrt{3} \):
\( 3st = st + 6s \)
Subtract \( st \) from both sides:
\( 2st = 6s \)
Divide both sides by \( 2s \) (assuming \( s \neq 0 \) since the van is moving):
\( t = \frac{6s}{2s} \)
\( t = 3 \) minutes.
Therefore, the van will take 3 more minutes to reach the tower.
In simple words: A man on a tower sees a van. The angle of depression changes from 30 to 60 degrees as the van moves towards the tower for 6 minutes. We use tangent formulas to relate the tower's height to the distances from the tower. By setting up two equations and solving them, we find out that the van will take 3 more minutes to reach the tower.

🎯 Exam Tip: When a vehicle moves towards or away from a base, always consider the speed and time to calculate the distance traveled. For angles of depression, use alternate interior angles to form angles inside the right triangles you draw.