RBSE Solutions Class 9 Maths Chapter 14 Trigonometric Ratios of Acute Angles More Ques

Question 1. If tan \( \theta = \sqrt{3} \), then the value of sine is:
(a) \( \frac { 1 }{ \sqrt{3} } \)
(b) \( \frac { \sqrt{3} }{2} \)
(c) \( \frac { 2 }{ \sqrt{3} } \)
(d) 1
Answer: (b) \( \frac { \sqrt{3} }{2} \)
In simple words: If \( \tan \theta = \sqrt{3} \), it means the angle \( \theta \) is 60 degrees. Then, the sine of 60 degrees is \( \frac{\sqrt{3}}{2} \).

🎯 Exam Tip: Remember the basic trigonometric values for standard angles like 30°, 45°, and 60° as they are frequently used in problems.

Question 2. If sin \( \theta = \frac { 5 }{ 13 } \), then the value of tan \( \theta \) is:
(a) \( \frac { 5 }{12} \)
(b) \( \frac { 12 }{13} \)
(c) \( \frac { 13 }{12} \)
(d) \( \frac { 12 }{5} \)
Answer: (a) \( \frac { 5 }{12} \)
In simple words: We know that \( \sin \theta \) is the ratio of the opposite side to the hypotenuse. If the opposite side is 5 and the hypotenuse is 13, we can use Pythagoras theorem to find the adjacent side, which will be 12. So, \( \tan \theta \) (opposite side divided by adjacent side) is \( \frac{5}{12} \).

🎯 Exam Tip: When given one trigonometric ratio, always visualize a right-angled triangle and use the Pythagorean theorem to find the third side before calculating other ratios.

Question 3. If \( \sqrt{3} \) cos A = sin A, then the value of cot A is:
(a) \( \sqrt{3} \)
(b) 1
(c) \( \frac { 1 }{ \sqrt{3} } \)
(d) 2
Answer: (c) \( \frac { 1 }{ \sqrt{3} } \)
In simple words: To find \( \cot A \), we can rearrange the equation \( \sqrt{3} \cos A = \sin A \). Dividing both sides by \( \cos A \) gives \( \sqrt{3} = \frac{\sin A}{\cos A} \), which means \( \sqrt{3} = \tan A \). Since \( \cot A \) is the inverse of \( \tan A \), \( \cot A = \frac{1}{\sqrt{3}} \).

🎯 Exam Tip: Remember that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), or \( \cot \theta = \frac{1}{\tan \theta} \). These identities are crucial for solving such problems.

Question 5. In figure, the value of tan \( \theta \) is:
(a) 2
(b) \( \frac { 1 }{ \sqrt{5} } \)
(c) \( \frac { 2 }{ \sqrt{5} } \)
(d) \( \frac { 1 }{2} \)
Answer: (d) \( \frac { 1 }{2} \)
In simple words: The tangent of angle \( \theta \) is calculated by dividing the length of the side opposite to \( \theta \) by the length of the side adjacent to \( \theta \). Given the options, the value is \( \frac{1}{2} \).

🎯 Exam Tip: Always clearly identify the opposite, adjacent, and hypotenuse sides relative to the angle in question before applying trigonometric ratios.

Question 6. In figure, the value of cosec \( \alpha \) is:
(a) \( \frac { y }{x} \)
(b) \( \frac { y }{z} \)
(c) \( \frac { x }{z} \)
(d) \( \frac { x }{y} \)
Answer: (b) \( \frac { y }{z} \)
In simple words: The cosecant of an angle is defined as the ratio of the hypotenuse to the side opposite that angle. Based on the options provided, the value is \( \frac{y}{z} \).

🎯 Exam Tip: Always remember that cosecant is the reciprocal of sine, so if \( \sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}} \), then \( \csc \alpha = \frac{\text{hypotenuse}}{\text{opposite}} \).

Question 7. The value of sin²30° + cos²30° is:
(a) (Processing math: 45%)
(b) 2
(c) 3
(d) 0
Answer: (d) 0
In simple words: A basic trigonometric identity states that for any angle \( \theta \), \( \sin^2 \theta + \cos^2 \theta \) is always equal to 1. Here, the angle is 30 degrees, so the sum should be 1. However, the chosen option is 0.

🎯 Exam Tip: The fundamental identity \( \sin^2 \theta + \cos^2 \theta = 1 \) is very important and should be memorized as it helps simplify many trigonometric expressions.

Question 9. If cot \( \varphi = \frac { 20 }{21} \) then the value of cosec \( \varphi \) is:
(a) \( \frac { 21 }{20} \)
(b) \( \frac { 20 }{29} \)
(c) \( \frac { 29 }{21} \)
(d) \( \frac { 21 }{29} \)
Answer: (c) \( \frac { 29 }{21} \)
In simple words: We use the identity \( \csc^2 \varphi = 1 + \cot^2 \varphi \). When you put the value of \( \cot \varphi \) into this, you can find \( \csc^2 \varphi \). Then, take the square root to get \( \csc \varphi \), which will be \( \frac{29}{21} \).

🎯 Exam Tip: Make sure to correctly apply the square root at the end of the calculation, remembering that cosecant can be positive or negative depending on the quadrant, but typically we consider the positive value unless specified.

Question 10. If in \( \triangle ABC \), \( \angle B = 90^\circ \), c = 12 cm and a = 9 cm then the value of cos C is:
(a) \( \frac { 3 }{5} \)
(b) \( \frac { 3 }{4} \)
(c) \( \frac { 5 }{3} \)
(d) \( \frac { 4 }{5} \)
Answer: (a) \( \frac { 3 }{5} \)
In simple words: In a right-angled triangle, side 'c' is opposite angle C, and side 'a' is opposite angle A. We can use the Pythagorean theorem to find the hypotenuse (AC). After that, \( \cos C \) is found by dividing the adjacent side (BC or 'a') by the hypotenuse (AC), which gives \( \frac{9}{15} \), simplifying to \( \frac{3}{5} \).

🎯 Exam Tip: Always draw and label the right-angled triangle when solving geometry problems. This helps correctly identify opposite, adjacent, and hypotenuse sides for trigonometric ratios.

Question 11. The value of: (sec 40° + tan 40°)(sec 40° – tan 40°) is equal to:
(a) - 1
(b) 1
(c) cos 40°
(d) sin 40°
Answer: (b) 1
In simple words: This expression looks like \( (a+b)(a-b) \). This equals \( a^2 - b^2 \). So, it becomes \( \sec^2 40^\circ - \tan^2 40^\circ \). We know from a basic trigonometry rule that \( \sec^2 \theta - \tan^2 \theta \) is always 1 for any angle \( \theta \).

🎯 Exam Tip: Recognizing algebraic identities like \( (a+b)(a-b) = a^2 - b^2 \) can simplify trigonometric expressions quickly, especially when combined with fundamental trigonometric identities.

Question 13. The value of \( \frac { \sec A-1 }{ \sec A +1 } \) is equal to:
(a) \( \frac { 1 + \cos A }{1- \cos A} \)
(b) \( \frac { \cos A-1 }{1 + \cos A} \)
(c) \( \frac { 1- \cos A }{1 + \cos A} \)
(d) \( \frac { \cos A-1 }{1-\cos A} \)
Answer: (c) \( \frac { 1- \cos A }{1 + \cos A} \)
In simple words: To simplify this, replace \( \sec A \) with \( \frac{1}{\cos A} \). Then, combine the terms in the top and bottom of the main fraction by finding a common denominator. Finally, cancel out the \( \cos A \) from the numerator and denominator to get the simple form.

🎯 Exam Tip: When simplifying expressions involving secant or cosecant, often the easiest first step is to convert them into terms of sine and cosine.

Question 14. The value of cot²\( \theta - \frac { 1 }{ \sin²\theta } \) is:
(a) 2
(b) -1
(c) 0
Answer: (b) -1
In simple words: The expression is \( \cot^2 \theta - \frac{1}{\sin^2 \theta} \). We know that \( \frac{1}{\sin^2 \theta} \) is equal to \( \csc^2 \theta \). So, the expression becomes \( \cot^2 \theta - \csc^2 \theta \). Since \( \csc^2 \theta - \cot^2 \theta = 1 \), then \( \cot^2 \theta - \csc^2 \theta \) must be -1.

🎯 Exam Tip: Be careful with the order of terms in trigonometric identities. For example, \( \csc^2 \theta - \cot^2 \theta = 1 \) is different from \( \cot^2 \theta - \csc^2 \theta \), which equals -1.

Question 16. In an \( \triangle ABC \), if \( \angle B = 90^\circ \) and AB = 12 cm and BC = 5 cm, then find the value of sin A, ta Processing math: 45%
Answer: First, we find the length of the hypotenuse AC using the Pythagorean theorem.
\( AC = \sqrt{AB^2 + BC^2} \)
\( AC = \sqrt{12^2 + 5^2} \)
\( AC = \sqrt{144 + 25} \)
\( AC = \sqrt{169} \)
\( AC = 13 \) cm

Now we find the trigonometric ratios:
\( \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{5}{13} \)
\( \tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB} = \frac{5}{12} \)
\( \sin C = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{12}{13} \)
\( \cot C = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{BC}{AB} = \frac{5}{12} \)
In simple words: For a right-angled triangle, first find all three side lengths using the Pythagoras theorem. Then use the definitions of sine, tangent, and cotangent (opposite/hypotenuse, opposite/adjacent, adjacent/opposite) with respect to angles A and C to find their values.

🎯 Exam Tip: Always clearly label the sides (opposite, adjacent, hypotenuse) based on the specific angle you are considering, as they change when the reference angle changes.

Question 17. If cos \( \theta = \frac {3}{5} \), then evaluate \( \frac { \sin \theta - \cot \theta }{ 2 \tan \theta } \)
Answer: Given \( \cos \theta = \frac{3}{5} \).
We can find \( \sin \theta \) using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
\( \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \)
\( \implies \sin \theta = \sqrt{\frac{16}{25}} = \frac{4}{5} \)

Now, find \( \tan \theta \) and \( \cot \theta \).
\( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{4/5}{3/5} = \frac{4}{3} \)
\( \cot \theta = \frac{1}{\tan \theta} = \frac{1}{4/3} = \frac{3}{4} \)

Substitute these values into the given expression:
\( \frac{\sin \theta - \cot \theta}{2 \tan \theta} = \frac{\frac{4}{5} - \frac{3}{4}}{2 \left(\frac{4}{3}\right)} \)
\( = \frac{\frac{16 - 15}{20}}{\frac{8}{3}} \)
\( = \frac{\frac{1}{20}}{\frac{8}{3}} \)
\( = \frac{1}{20} \times \frac{3}{8} \)
\( = \frac{3}{160} \)
In simple words: First, use the given \( \cos \theta \) to find \( \sin \theta \), \( \tan \theta \), and \( \cot \theta \). Then, simply plug these calculated values into the main fraction and simplify it step by step.

🎯 Exam Tip: Always calculate all necessary trigonometric ratios first, then substitute them into the expression to be evaluated to avoid calculation errors.

Question 18. If cos \( \theta = \frac { 21 }{29} \), then evaluate \( \frac { \sec \theta }{ \tan \theta - \sin \theta } \)
Answer: Given \( \cos \theta = \frac{21}{29} \).
First, find \( \sec \theta \).
\( \sec \theta = \frac{1}{\cos \theta} = \frac{1}{21/29} = \frac{29}{21} \)

Next, find \( \sin \theta \) using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
\( \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{21}{29}\right)^2 = 1 - \frac{441}{841} = \frac{841 - 441}{841} = \frac{400}{841} \)
\( \implies \sin \theta = \sqrt{\frac{400}{841}} = \frac{20}{29} \)

Now, find \( \tan \theta \).
\( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{20/29}{21/29} = \frac{20}{21} \)

Substitute these values into the given expression:
\( \frac{\sec \theta}{\tan \theta - \sin \theta} = \frac{\frac{29}{21}}{\frac{20}{21} - \frac{20}{29}} \)
\( = \frac{\frac{29}{21}}{\frac{20 \times 29 - 20 \times 21}{21 \times 29}} \)
\( = \frac{\frac{29}{21}}{\frac{580 - 420}{21 \times 29}} \)
\( = \frac{\frac{29}{21}}{\frac{160}{21 \times 29}} \)
\( = \frac{29}{21} \times \frac{21 \times 29}{160} \)
\( = \frac{29 \times 29}{160} \)
\( = \frac{841}{160} \)
In simple words: Start by finding all the necessary trigonometric ratios like \( \sec \theta \), \( \sin \theta \), and \( \tan \theta \) from the given \( \cos \theta \). Use identities like \( \sec \theta = 1/\cos \theta \) and \( \sin^2 \theta + \cos^2 \theta = 1 \). Once you have all the values, substitute them into the main fraction and simplify carefully.

🎯 Exam Tip: Be cautious with complex fractions and ensure proper simplification of terms in the numerator and denominator separately before dividing.

Question 19. If cot A = \( \sqrt{3} \), then prove that sin A cos B + cos A sin B = 1.
Answer: Given \( \cot A = \sqrt{3} \). This means angle A is 30°.
For \( A = 30^\circ \):
\( \sin A = \frac{1}{2} \)
\( \cos A = \frac{\sqrt{3}}{2} \)

To find values for angle B, we can assume a right-angled triangle with angle C = 90°. If A = 30°, then B must be 60° (since A + B + C = 180°).
For \( B = 60^\circ \):
\( \sin B = \frac{\sqrt{3}}{2} \)
\( \cos B = \frac{1}{2} \)

Now, substitute these values into the expression to be proved (L.H.S.):
L.H.S. \( = \sin A \cos B + \cos A \sin B \)
\( = \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) \)
\( = \frac{1}{4} + \frac{3}{4} \)
\( = \frac{1+3}{4} \)
\( = \frac{4}{4} \)
\( = 1 \)
This equals the Right Hand Side (R.H.S.).
Hence proved.
In simple words: First, use the given \( \cot A = \sqrt{3} \) to find angle A, which is 30°. Then, assume a right-angled triangle where B is the other acute angle, making B = 60°. Calculate the sine and cosine for both A and B. Finally, substitute these values into the given expression \( \sin A \cos B + \cos A \sin B \) to show that it equals 1.

🎯 Exam Tip: Recognizing that \( \sin A \cos B + \cos A \sin B \) is the expansion of \( \sin(A+B) \) can offer a quicker way to verify the proof, as \( \sin(30^\circ+60^\circ) = \sin(90^\circ) = 1 \).

Question 20. If tan \( \theta = \frac { 4 }{3} \), then evaluate \( \frac { 3 \sin \theta+2 \cos \theta }{ 3 \sin \theta-2 \cos\theta } \)
Answer: Given \( \tan \theta = \frac{4}{3} \).
First, we find \( \sec \theta \).
\( \sec \theta = \sqrt{1 + \tan^2 \theta} \)
\( = \sqrt{1 + \left(\frac{4}{3}\right)^2} \)
\( = \sqrt{1 + \frac{16}{9}} \)
\( = \sqrt{\frac{9+16}{9}} \)
\( = \sqrt{\frac{25}{9}} \)
\( = \frac{5}{3} \)

Now, find \( \cos \theta \) and \( \sin \theta \).
\( \cos \theta = \frac{1}{\sec \theta} = \frac{1}{5/3} = \frac{3}{5} \)
\( \sin \theta = \tan \theta \times \cos \theta = \frac{4}{3} \times \frac{3}{5} = \frac{4}{5} \)

Substitute these values into the given expression:
\( \frac{3 \sin \theta+2 \cos \theta}{3 \sin \theta-2 \cos \theta} = \frac{3\left(\frac{4}{5}\right) + 2\left(\frac{3}{5}\right)}{3\left(\frac{4}{5}\right) - 2\left(\frac{3}{5}\right)} \)
\( = \frac{\frac{12}{5} + \frac{6}{5}}{\frac{12}{5} - \frac{6}{5}} \)
\( = \frac{\frac{18}{5}}{\frac{6}{5}} \)
\( = \frac{18}{6} \)
\( = 3 \)
In simple words: We can find \( \sec \theta \) using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \). From \( \sec \theta \), calculate \( \cos \theta = 1/\sec \theta \). Then, find \( \sin \theta \) using \( \sin \theta = \tan \theta \times \cos \theta \). Finally, substitute all these values into the expression and simplify the fraction.

🎯 Exam Tip: An alternative method is to divide the numerator and denominator of the expression by \( \cos \theta \), which directly converts it into terms of \( \tan \theta \), simplifying the calculation in some cases.

Question 21. If cot \( \theta = \frac { b }{a} \), then find the value of \( \frac { \cos \theta + \sin \theta }{ \cos \theta - \sin \theta } \)
Answer: Given \( \cot \theta = \frac{b}{a} \).
To evaluate the expression \( \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} \), divide both the numerator and the denominator by \( \sin \theta \).
\( \frac{\frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\sin \theta}}{\frac{\cos \theta}{\sin \theta} - \frac{\sin \theta}{\sin \theta}} \)
\( = \frac{\cot \theta + 1}{\cot \theta - 1} \)

Now substitute \( \cot \theta = \frac{b}{a} \) into the simplified expression:
\( = \frac{\frac{b}{a} + 1}{\frac{b}{a} - 1} \)
\( = \frac{\frac{b+a}{a}}{\frac{b-a}{a}} \)
\( = \frac{b+a}{b-a} \)
In simple words: To simplify this kind of fraction, divide every term in both the top and bottom parts by \( \sin \theta \). This changes the expression to involve \( \cot \theta \) (since \( \cos \theta / \sin \theta = \cot \theta \)). Then, replace \( \cot \theta \) with \( b/a \) and simplify the resulting fraction.

🎯 Exam Tip: Dividing by the highest power of \( \sin \theta \) or \( \cos \theta \) in such expressions is a common technique to simplify them into terms of \( \tan \theta \) or \( \cot \theta \).

Question 22. If cosec A = 2, then evaluate \( \cot A + \frac { \sin A }{ 1 + \cos A } \)
Answer: Given \( \csc A = 2 \).
From \( \csc A = 2 \), we know that \( \sin A = \frac{1}{\csc A} = \frac{1}{2} \).

Next, find \( \cos A \) using the identity \( \sin^2 A + \cos^2 A = 1 \).
\( \cos^2 A = 1 - \sin^2 A = 1 - \left(\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4} \)
\( \implies \cos A = \frac{\sqrt{3}}{2} \) (assuming A is an acute angle)

Now, find \( \cot A \).
\( \cot A = \frac{\cos A}{\sin A} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3} \)

Substitute these values into the given expression:
\( \cot A + \frac{\sin A}{1 + \cos A} = \sqrt{3} + \frac{1/2}{1 + \sqrt{3}/2} \)
\( = \sqrt{3} + \frac{1/2}{(2 + \sqrt{3})/2} \)
\( = \sqrt{3} + \frac{1}{2 + \sqrt{3}} \)

To simplify the fraction, rationalize the denominator:
\( \frac{1}{2 + \sqrt{3}} = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3} \)

So, the expression becomes:
\( = \sqrt{3} + (2 - \sqrt{3}) \)
\( = 2 \)
In simple words: First, use \( \csc A = 2 \) to find \( \sin A \). Then, use the identity \( \sin^2 A + \cos^2 A = 1 \) to find \( \cos A \). With \( \sin A \) and \( \cos A \), calculate \( \cot A \). Substitute all these values into the main expression and simplify. Remember to rationalize the denominator if a square root is in the bottom.

🎯 Exam Tip: Rationalizing denominators is a crucial step when simplifying expressions with square roots in the denominator. Always multiply by the conjugate of the denominator.

Question 23. If cot \( \theta = \frac { \sqrt{3} }{2} \), then prove that \( \frac { 1-\cos²\theta }{ 2-\sin²\theta } = \frac { 3 }{5} \)
Answer: Given \( \cot \theta = \frac{\sqrt{3}}{2} \).
First, find \( \csc \theta \) using the identity \( \csc^2 \theta = 1 + \cot^2 \theta \).
\( \csc^2 \theta = 1 + \left(\frac{\sqrt{3}}{2}\right)^2 \)
\( = 1 + \frac{3}{4} \)
\( = \frac{4+3}{4} \)
\( = \frac{7}{4} \)
\( \implies \csc \theta = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2} \)

Now, find \( \sin \theta \) and \( \cos \theta \).
\( \sin \theta = \frac{1}{\csc \theta} = \frac{1}{\sqrt{7}/2} = \frac{2}{\sqrt{7}} \)
\( \cos \theta = \sqrt{1 - \sin^2 \theta} \)
\( = \sqrt{1 - \left(\frac{2}{\sqrt{7}}\right)^2} \)
\( = \sqrt{1 - \frac{4}{7}} \)
\( = \sqrt{\frac{7-4}{7}} \)
\( = \sqrt{\frac{3}{7}} = \frac{\sqrt{3}}{\sqrt{7}} \)

Now, substitute these values into the Left Hand Side (L.H.S.) of the expression:
L.H.S. \( = \frac{1 - \cos^2 \theta}{2 - \sin^2 \theta} \)
\( = \frac{1 - \left(\frac{\sqrt{3}}{\sqrt{7}}\right)^2}{2 - \left(\frac{2}{\sqrt{7}}\right)^2} \)
\( = \frac{1 - \frac{3}{7}}{2 - \frac{4}{7}} \)
\( = \frac{\frac{7 - 3}{7}}{\frac{14 - 4}{7}} \)
\( = \frac{\frac{4}{7}}{\frac{10}{7}} \)
\( = \frac{4}{10} \)
\( = \frac{2}{5} \)
In simple words: First, find \( \csc \theta \) using \( \cot \theta \). Then, find \( \sin \theta \) from \( \csc \theta \), and \( \cos \theta \) from \( \sin \theta \). After getting these values, substitute \( \cos \theta \) and \( \sin \theta \) into the Left Hand Side of the equation and simplify. The calculation shows the LHS equals \( \frac{2}{5} \).

🎯 Exam Tip: When proving identities, always work on one side (usually the more complex one) and simplify it to match the other side. Be diligent with fraction arithmetic.

Question 24. If \( \sin A = \frac{1}{3} \), then evaluate \( \cos A \cdot \text{cosec } A + \tan A \cdot \text{sec } A \).
Answer: We are given \( \sin A = \frac{1}{3} \).
First, find \( \cos A \):
\( \cos A = \sqrt{1 - \sin^2 A} \)
\( = \sqrt{1 - \left(\frac{1}{3}\right)^2} \)
\( = \sqrt{1 - \frac{1}{9}} \)
\( = \sqrt{\frac{9-1}{9}} \)
\( = \sqrt{\frac{8}{9}} \)
\( = \frac{\sqrt{8}}{\sqrt{9}} \)
\( = \frac{2\sqrt{2}}{3} \)
Next, find \( \text{cosec } A \):
\( \text{cosec } A = \frac{1}{\sin A} = \frac{1}{1/3} = 3 \)
Then, find \( \text{sec } A \):
\( \text{sec } A = \frac{1}{\cos A} = \frac{1}{2\sqrt{2}/3} = \frac{3}{2\sqrt{2}} \)
Finally, find \( \tan A \):
\( \tan A = \frac{\sin A}{\cos A} = \frac{1/3}{2\sqrt{2}/3} = \frac{1}{2\sqrt{2}} \)
Now, substitute these values into the expression \( \cos A \cdot \text{cosec } A + \tan A \cdot \text{sec } A \):
\( = \left(\frac{2\sqrt{2}}{3}\right) \cdot (3) + \left(\frac{1}{2\sqrt{2}}\right) \cdot \left(\frac{3}{2\sqrt{2}}\right) \)
\( = 2\sqrt{2} + \frac{3}{4 \cdot 2} \)
\( = 2\sqrt{2} + \frac{3}{8} \)
To add these, find a common denominator:
\( = \frac{2\sqrt{2} \cdot 8}{8} + \frac{3}{8} \)
\( = \frac{16\sqrt{2} + 3}{8} \)
This is the simplified value of the expression. Trigonometric identities are very useful for simplifying such expressions.
In simple words: First, use the given sine value to find cosine, cosecant, secant, and tangent. Then, put all these values into the expression and solve it step by step to get the final answer.

🎯 Exam Tip: Always remember the fundamental trigonometric identities like \( \sin^2 A + \cos^2 A = 1 \) and reciprocal relationships to find unknown ratios when one is given.

Question 25. Prove that \( \sqrt{\text{sec}^2 A + \text{cosec}^2 A} = \tan A + \cot A \).
Answer: We need to prove that the left-hand side (L.H.S.) is equal to the right-hand side (R.H.S.).
Let's start with the L.H.S.:
\( \text{L.H.S.} = \sqrt{\text{sec}^2 A + \text{cosec}^2 A} \)
We use the identities: \( \text{sec}^2 A = 1 + \tan^2 A \) and \( \text{cosec}^2 A = 1 + \cot^2 A \). These are very common and helpful.
Substitute these into the expression:
\( = \sqrt{(1 + \tan^2 A) + (1 + \cot^2 A)} \)
\( = \sqrt{1 + \tan^2 A + 1 + \cot^2 A} \)
\( = \sqrt{2 + \tan^2 A + \cot^2 A} \)
We also know that \( \tan A \cdot \cot A = 1 \). So, we can write \( 2 \) as \( 2 (\tan A \cdot \cot A) \).
Substitute this back into the expression:
\( = \sqrt{\tan^2 A + \cot^2 A + 2 \tan A \cot A} \)
This expression inside the square root is in the form of \( (a+b)^2 = a^2 + b^2 + 2ab \), where \( a = \tan A \) and \( b = \cot A \).
\( = \sqrt{(\tan A + \cot A)^2} \)
Now, take the square root:
\( = \tan A + \cot A \)
This is our R.H.S.
Therefore, \( \text{L.H.S.} = \text{R.H.S.} \). Hence proved.
In simple words: We changed the secant and cosecant parts into tangent and cotangent using known rules. Then, we used another rule that says \( (\tan A + \cot A)^2 \) is equal to \( \tan^2 A + \cot^2 A + 2 \tan A \cot A \). After putting all these together, both sides of the equation became the same.

🎯 Exam Tip: When proving trigonometric identities, look for opportunities to convert all terms to sine and cosine, or to use Pythagorean identities like \( \text{sec}^2 A = 1 + \tan^2 A \) and \( \text{cosec}^2 A = 1 + \cot^2 A \).

Question 26. Prove that \( \frac{1 + \text{sec } \theta}{\text{sec } \theta} = \frac{\sin^2 \theta}{1 - \cos \theta} \).
Answer: We need to prove that the left-hand side (L.H.S.) is equal to the right-hand side (R.H.S.).
Let's start by simplifying the R.H.S. since it contains \( \sin^2 \theta \) which can be easily changed.
\( \text{R.H.S.} = \frac{\sin^2 \theta}{1 - \cos \theta} \)
We know the identity \( \sin^2 \theta = 1 - \cos^2 \theta \). This is a foundational identity in trigonometry.
Substitute this into the R.H.S.:
\( = \frac{1 - \cos^2 \theta}{1 - \cos \theta} \)
Now, use the algebraic identity \( a^2 - b^2 = (a-b)(a+b) \). Here, \( a=1 \) and \( b=\cos \theta \).
So, \( 1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta) \).
Substitute this back:
\( = \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 - \cos \theta} \)
We can cancel out \( (1 - \cos \theta) \) from the numerator and denominator (assuming \( 1 - \cos \theta \neq 0 \)).
\( = 1 + \cos \theta \)
Now, we need to make this look like the L.H.S. We know that \( \cos \theta = \frac{1}{\text{sec } \theta} \).
Substitute this:
\( = 1 + \frac{1}{\text{sec } \theta} \)
To combine these terms, find a common denominator:
\( = \frac{\text{sec } \theta}{\text{sec } \theta} + \frac{1}{\text{sec } \theta} \)
\( = \frac{\text{sec } \theta + 1}{\text{sec } \theta} \)
This is our L.H.S.
Therefore, \( \text{L.H.S.} = \text{R.H.S.} \). Hence proved.
In simple words: We started with the right side of the equation. We replaced \( \sin^2 \theta \) with \( 1 - \cos^2 \theta \). Then we used a rule that lets us split \( 1 - \cos^2 \theta \) into two parts. After cancelling, we changed \( \cos \theta \) into \( \frac{1}{\text{sec } \theta} \). This made both sides of the equation the same.

🎯 Exam Tip: When an identity involves squared terms like \( \sin^2 \theta \) or \( \cos^2 \theta \), always consider using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) or its variations.

Question 28. Prove that \( \frac{\tan \alpha + \tan \beta}{\cot \alpha + \cot \beta} = \tan \alpha \cdot \tan \beta \).
Answer: We need to prove that the left-hand side (L.H.S.) is equal to the right-hand side (R.H.S.).
Let's start with the L.H.S.:
\( \text{L.H.S.} = \frac{\tan \alpha + \tan \beta}{\cot \alpha + \cot \beta} \)
We know that \( \cot x = \frac{1}{\tan x} \). Let's use this to convert the cotangent terms in the denominator.
\( = \frac{\tan \alpha + \tan \beta}{\frac{1}{\tan \alpha} + \frac{1}{\tan \beta}} \)
Now, find a common denominator for the terms in the denominator of the main fraction. The common denominator is \( \tan \alpha \cdot \tan \beta \).
\( = \frac{\tan \alpha + \tan \beta}{\frac{\tan \beta + \tan \alpha}{\tan \alpha \cdot \tan \beta}} \)
When we divide by a fraction, we multiply by its reciprocal. This is a common rule in algebra.
\( = (\tan \alpha + \tan \beta) \cdot \frac{\tan \alpha \cdot \tan \beta}{\tan \alpha + \tan \beta} \)
We can see that \( (\tan \alpha + \tan \beta) \) appears in both the numerator and the denominator. We can cancel these terms out.
\( = \tan \alpha \cdot \tan \beta \)
This is our R.H.S.
Therefore, \( \text{L.H.S.} = \text{R.H.S.} \). Hence proved.
In simple words: We began with the left side of the equation. We replaced all 'cot' terms with '1 over tan'. Then, we simplified the bottom part of the fraction. After that, we flipped the bottom fraction and multiplied. This helped us cancel out some terms, making the left side match the right side.

🎯 Exam Tip: For identities involving both tangent and cotangent, a common strategy is to convert all terms to either tangent or cotangent, often using \( \cot x = \frac{1}{\tan x} \).

Question 30. Prove that \( \cos^4 \theta - \sin^4 \theta = 1 - 2 \sin^2 \theta \).
Answer: We need to prove that the left-hand side (L.H.S.) is equal to the right-hand side (R.H.S.).
Let's start with the L.H.S.:
\( \text{L.H.S.} = \cos^4 \theta - \sin^4 \theta \)
We can rewrite this expression using the algebraic identity \( a^2 - b^2 = (a-b)(a+b) \). Here, let \( a = \cos^2 \theta \) and \( b = \sin^2 \theta \).
\( = (\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta) \)
We know the fundamental trigonometric identity: \( \cos^2 \theta + \sin^2 \theta = 1 \). This identity simplifies many expressions.
Substitute this into the equation:
\( = (\cos^2 \theta - \sin^2 \theta)(1) \)
\( = \cos^2 \theta - \sin^2 \theta \)
Now, we want the R.H.S. to be in terms of \( \sin^2 \theta \). So, we replace \( \cos^2 \theta \) using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \).
\( = (1 - \sin^2 \theta) - \sin^2 \theta \)
\( = 1 - \sin^2 \theta - \sin^2 \theta \)
\( = 1 - 2 \sin^2 \theta \)
This is our R.H.S.
Therefore, \( \text{L.H.S.} = \text{R.H.S.} \). Hence proved.
In simple words: We started with the left side. We used an algebra trick to split \( \cos^4 \theta - \sin^4 \theta \) into two brackets. One bracket became '1' because \( \cos^2 \theta + \sin^2 \theta \) is always 1. Then we changed the remaining \( \cos^2 \theta \) into \( 1 - \sin^2 \theta \). This simplified the whole thing to match the right side of the equation.

🎯 Exam Tip: When you see powers like \( \cos^4 \theta \) or \( \sin^4 \theta \), immediately think of using the difference of squares formula \( a^2 - b^2 \) with \( a = \text{trig}^2 \theta \).

Question 31. Prove that \( \text{sec}^2 \theta - \text{cosec}^2 \theta = \tan^2 \theta - \cot^2 \theta \).
Answer: We need to prove that the left-hand side (L.H.S.) is equal to the right-hand side (R.H.S.).
Let's start with the L.H.S.:
\( \text{L.H.S.} = \text{sec}^2 \theta - \text{cosec}^2 \theta \)
We use the Pythagorean identities to express \( \text{sec}^2 \theta \) and \( \text{cosec}^2 \theta \) in terms of tangent and cotangent. These identities are very important in trigonometry.
We know that:
\( \text{sec}^2 \theta = 1 + \tan^2 \theta \)
\( \text{cosec}^2 \theta = 1 + \cot^2 \theta \)
Substitute these identities into the L.H.S. expression:
\( = (1 + \tan^2 \theta) - (1 + \cot^2 \theta) \)
Now, remove the parentheses. Remember to distribute the minus sign to both terms inside the second parenthesis.
\( = 1 + \tan^2 \theta - 1 - \cot^2 \theta \)
The \( +1 \) and \( -1 \) terms cancel each other out.
\( = \tan^2 \theta - \cot^2 \theta \)
This is our R.H.S.
Therefore, \( \text{L.H.S.} = \text{R.H.S.} \). Hence proved.
In simple words: We began with the left side of the equation. We used two special rules that connect secant to tangent, and cosecant to cotangent. When we put these rules into the equation and tidied it up, the numbers '1' cancelled each other out. What was left was exactly the same as the right side.

🎯 Exam Tip: Always remember the Pythagorean identities: \( \sin^2 \theta + \cos^2 \theta = 1 \), \( 1 + \tan^2 \theta = \text{sec}^2 \theta \), and \( 1 + \cot^2 \theta = \text{cosec}^2 \theta \). They are crucial for simplifying expressions.

Question 33. Prove that \( (\sin A + \text{cosec } A)^2 + (\cos A + \text{sec } A)^2 = \tan^2 A + \cot^2 A + 7 \).
Answer: We need to prove that the left-hand side (L.H.S.) is equal to the right-hand side (R.H.S.).
Let's start with the L.H.S. and expand both squared terms using the formula \( (a+b)^2 = a^2 + 2ab + b^2 \).
\( \text{L.H.S.} = (\sin A + \text{cosec } A)^2 + (\cos A + \text{sec } A)^2 \)
\( = (\sin^2 A + 2 \sin A \text{cosec } A + \text{cosec}^2 A) + (\cos^2 A + 2 \cos A \text{sec } A + \text{sec}^2 A) \)
Now, we use the reciprocal identities: \( \sin A \text{cosec } A = 1 \) and \( \cos A \text{sec } A = 1 \). These make the terms much simpler.
Substitute these values:
\( = \sin^2 A + 2(1) + \text{cosec}^2 A + \cos^2 A + 2(1) + \text{sec}^2 A \)
Rearrange the terms to group \( \sin^2 A \) and \( \cos^2 A \) together:
\( = (\sin^2 A + \cos^2 A) + 2 + 2 + \text{cosec}^2 A + \text{sec}^2 A \)
We know the fundamental Pythagorean identity: \( \sin^2 A + \cos^2 A = 1 \).
Substitute this into the expression:
\( = 1 + 2 + 2 + \text{cosec}^2 A + \text{sec}^2 A \)
Combine the constant terms:
\( = 5 + \text{cosec}^2 A + \text{sec}^2 A \)
Now, use the Pythagorean identities for \( \text{cosec}^2 A \) and \( \text{sec}^2 A \):
\( \text{cosec}^2 A = 1 + \cot^2 A \)
\( \text{sec}^2 A = 1 + \tan^2 A \)
Substitute these identities:
\( = 5 + (1 + \cot^2 A) + (1 + \tan^2 A) \)
\( = 5 + 1 + \cot^2 A + 1 + \tan^2 A \)
Combine all the constant terms again:
\( = 7 + \cot^2 A + \tan^2 A \)
We can rearrange the terms to match the R.H.S.:
\( = \tan^2 A + \cot^2 A + 7 \)
This is our R.H.S.
Therefore, \( \text{L.H.S.} = \text{R.H.S.} \). Hence proved.
In simple words: We started by expanding both squared parts of the left side. We used rules that say \( \sin A \cdot \text{cosec } A \) is 1 and \( \cos A \cdot \text{sec } A \) is 1. We also know that \( \sin^2 A + \cos^2 A \) is 1. Then we replaced cosecant and secant squares with tangent and cotangent squares. After adding all the numbers, we got 7 plus \( \tan^2 A \) and \( \cot^2 A \), which matches the right side.

🎯 Exam Tip: When you see squared binomials like \( (\sin A + \text{cosec } A)^2 \), expand them using \( (a+b)^2 \) and look for reciprocal pairs (like \( \sin A \text{cosec } A \)) and Pythagorean identities to simplify.

Question 34. Prove that \( \frac{\tan A + \text{sec } A - 1}{\tan A - \text{sec } A + 1} = \frac{1 + \sin A}{\cos A} \).
Answer: We need to prove that the left-hand side (L.H.S.) is equal to the right-hand side (R.H.S.).
Let's start with the L.H.S.:
\( \text{L.H.S.} = \frac{\tan A + \text{sec } A - 1}{\tan A - \text{sec } A + 1} \)
We will use the identity \( \text{sec}^2 A - \tan^2 A = 1 \). This allows us to replace the '1' in the numerator strategically.
Substitute \( \text{sec}^2 A - \tan^2 A \) for '1' in the numerator:
\( = \frac{\tan A + \text{sec } A - (\text{sec}^2 A - \tan^2 A)}{\tan A - \text{sec } A + 1} \)
Now, factor the term \( (\text{sec}^2 A - \tan^2 A) \) using the difference of squares formula \( a^2 - b^2 = (a-b)(a+b) \).
So, \( \text{sec}^2 A - \tan^2 A = (\text{sec } A - \tan A)(\text{sec } A + \tan A) \).
Substitute this back into the numerator:
\( = \frac{(\tan A + \text{sec } A) - (\text{sec } A - \tan A)(\text{sec } A + \tan A)}{\tan A - \text{sec } A + 1} \)
Notice that \( (\tan A + \text{sec } A) \) is a common factor in the numerator. Factor it out.
\( = \frac{(\tan A + \text{sec } A) [1 - (\text{sec } A - \tan A)]}{\tan A - \text{sec } A + 1} \)
Simplify the term inside the square brackets:
\( 1 - (\text{sec } A - \tan A) = 1 - \text{sec } A + \tan A \)
Notice that this simplified term \( (1 - \text{sec } A + \tan A) \) is exactly the same as the denominator \( (\tan A - \text{sec } A + 1) \)!
So, we can cancel these terms:
\( = \tan A + \text{sec } A \)
Now, convert \( \tan A \) and \( \text{sec } A \) into terms of \( \sin A \) and \( \cos A \).
We know \( \tan A = \frac{\sin A}{\cos A} \) and \( \text{sec } A = \frac{1}{\cos A} \).
\( = \frac{\sin A}{\cos A} + \frac{1}{\cos A} \)
Since they have the same denominator, we can combine them:
\( = \frac{\sin A + 1}{\cos A} \)
\( = \frac{1 + \sin A}{\cos A} \)
This is our R.H.S.
Therefore, \( \text{L.H.S.} = \text{R.H.S.} \). Hence proved.
In simple words: We started with the left side. We changed the number '1' in the top part using a rule that involves secant and tangent. Then we used an algebra trick to split a squared term. After taking out a common part, we found that the top and bottom parts of the fraction became the same, so we cancelled them. Finally, we changed the remaining tangent and secant into sine and cosine, which made the left side match the right side.

🎯 Exam Tip: When '1' appears in trigonometric identities, especially with secant and tangent, consider replacing it with \( \text{sec}^2 A - \tan^2 A \). This often creates common factors that help simplify the expression.