Samacheer Kalvi Class 9 Maths Solutions Chapter 6 Trigonometry More Ques

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Additional Questions

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Additional Questions

I. Choose the Correct Answer

Question 1. The value of \( \text{cosec}^2 60^{\circ} - 1 \) is equal to ........
(a) \( \cos^2 60^{\circ} \)
(b) \( \cot^2 60^{\circ} \)
(c) \( \cos^2 60 \)
(d) \( \tan 60 \)
Answer: (b) \( \cot^2 60^{\circ} \)
In simple words: The value of cosecant squared of 60 degrees minus 1 is the same as cotangent squared of 60 degrees. This is based on a basic trigonometric identity.

🎯 Exam Tip: Remember the Pythagorean identity \( 1 + \cot^2 \theta = \text{cosec}^2 \theta \), which can be rearranged to \( \text{cosec}^2 \theta - 1 = \cot^2 \theta \). Using this identity is key for such problems.

Question 2. The value of \( \cos 60^{\circ} \cos 30^{\circ} - \sin 60^{\circ} \sin 30^{\circ} \) is equal is ........
(a) \( \text{cosec } 90^{\circ} \)
(b) \( \tan 90^{\circ} \)
(c) \( \sin 30^{\circ} + \cos 30^{\circ} \)
(d) \( \cos 90^{\circ} \)
Answer: (d) \( \cos 90^{\circ} \)
In simple words: This expression is a trigonometric formula for \( \cos(A+B) \). When you calculate it, the result is the same as the cosine of 90 degrees, which is 0.

🎯 Exam Tip: Recognize the sum formula for cosine: \( \cos(A+B) = \cos A \cos B - \sin A \sin B \). Here, \( A = 60^{\circ} \) and \( B = 30^{\circ} \), so the expression simplifies to \( \cos(60^{\circ} + 30^{\circ}) = \cos 90^{\circ} \).

Question 3. The value of \( \frac{\sin 57^{\circ}}{\cos 33^{\circ}} \) is ........
(a) \( \cot 63^{\circ} \)
(b) \( \tan 27^{\circ} \)
(c) 1
(d) 0
Answer: (c) 1
In simple words: Since 57 degrees and 33 degrees add up to 90 degrees, sine of 57 degrees is the same as cosine of 33 degrees. So, dividing them gives 1.

🎯 Exam Tip: Remember the complementary angle identity: \( \sin \theta = \cos (90^{\circ} - \theta) \). Here, \( \sin 57^{\circ} = \cos (90^{\circ} - 57^{\circ}) = \cos 33^{\circ} \). This makes the fraction equal to 1.

Question 4. If \( 3 \text{ cosec } 36^{\circ} = \sec 54^{\circ} \) then the value of x is ........
(a) 0
(b) 1
(c) \( \frac{1}{3} \)
(d) \( \frac{3}{4} \)
Answer: (c) \( \frac{1}{3} \)
In simple words: We know that cosecant of an angle is the same as secant of its complementary angle. So, \( \text{cosec } 36^{\circ} \) is equal to \( \sec 54^{\circ} \). If we consider the equation to be \( 3x \text{ cosec } 36^{\circ} = \sec 54^{\circ} \), then by substituting, we find \( x \) to be \( \frac{1}{3} \).

🎯 Exam Tip: Apply the complementary angle identity \( \text{cosec } \theta = \sec (90^{\circ} - \theta) \). In this case, \( \text{cosec } 36^{\circ} = \sec (90^{\circ} - 36^{\circ}) = \sec 54^{\circ} \). This identity helps simplify the equation significantly. Make sure to clearly identify what needs to be solved for.

Question 5. If \( \cos A \cos 30^{\circ} = \frac{\sqrt{3}}{4} \), then the measures of A is ........
(a) \( 90^{\circ} \)
(b) \( 60^{\circ} \)
(c) \( 45^{\circ} \)
(d) \( 30^{\circ} \)
Answer: (b) \( 60^{\circ} \)
In simple words: We are given an equation with cosine of angle A and cosine of 30 degrees. By knowing the value of \( \cos 30^{\circ} \), we can find \( \cos A \), which then tells us the angle A.

🎯 Exam Tip: Substitute the known value of \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \) into the equation. This helps isolate \( \cos A \), allowing you to find the value of angle A by looking up standard trigonometric values.

II. Answer the Following Question

Question 1. Given Sec \( \theta = \frac{13}{12} \). Calculate all other trigonometric ratios.
Answer:
In the right triangle ABC:
We are given \( \sec \theta = \frac{13}{12} \). We know that \( \sec \theta = \frac{\text{hypotenuse}}{\text{adjacent side}} \).
So, if \( \theta \) is at angle A, then hypotenuse AC = 13 and adjacent side AB = 12.
Using the Pythagorean theorem, \( \text{BC}^2 = \text{AC}^2 - \text{AB}^2 \).
\( \text{BC}^2 = 13^2 - 12^2 \)
\( = 169 - 144 \)
\( = 25 \)
\( \implies \text{BC} = \sqrt{25} \)
\( = 5 \)
Now we can calculate all other trigonometric ratios:
\( \sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{\text{BC}}{\text{AC}} = \frac{5}{13} \)
\( \cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{\text{AB}}{\text{AC}} = \frac{12}{13} \)
\( \tan \theta = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{\text{BC}}{\text{AB}} = \frac{5}{12} \)
\( \text{cosec } \theta = \frac{\text{hypotenuse}}{\text{opposite side}} = \frac{\text{AC}}{\text{BC}} = \frac{13}{5} \)
\( \cot \theta = \frac{\text{adjacent side}}{\text{opposite side}} = \frac{\text{AB}}{\text{BC}} = \frac{12}{5} \)
In simple words: First, use the given secant value to find the sides of a right-angled triangle. Since secant is hypotenuse over adjacent, we get two sides. Then, use Pythagoras theorem to find the third side. Once all three sides are known, you can write down all six trigonometric ratios easily.

🎯 Exam Tip: Always draw a right-angled triangle first, label its sides correctly based on the given ratio and angle, and then use the Pythagorean theorem to find the missing side before calculating other ratios.

Question 2. If \( 3 \cot A = 4 \) check weather \( \frac{1- \tan^2 A}{1+ \tan^2 A} = \cos^2 A - \sin^2 A \) or not?
Answer:
Given \( 3 \cot A = 4 \), which means \( \cot A = \frac{4}{3} \).
In a right triangle ABC, let B be the right angle.
If \( \cot A = \frac{\text{adjacent side}}{\text{opposite side}} \), then AB = 4 and BC = 3.
Using the Pythagorean theorem to find the hypotenuse AC:
\( \text{AC}^2 = \text{AB}^2 + \text{BC}^2 \)
\( = 4^2 + 3^2 \)
\( = 16 + 9 \)
\( = 25 \)
\( \implies \text{AC} = \sqrt{25} \)
\( = 5 \)
Now we find the values of \( \tan A, \cos A, \sin A \):
\( \tan A = \frac{1}{\cot A} = \frac{3}{4} \)
\( \cos A = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{\text{AB}}{\text{AC}} = \frac{4}{5} \)
\( \sin A = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{\text{BC}}{\text{AC}} = \frac{3}{5} \)
Now, let's evaluate the Left Hand Side (L.H.S.):
\( \text{L.H.S.} = \frac{1- \tan^2 A}{1+ \tan^2 A} \)
\( = \frac{1 - \left(\frac{3}{4}\right)^2}{1 + \left(\frac{3}{4}\right)^2} \)
\( = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} \)
\( = \frac{\frac{16-9}{16}}{\frac{16+9}{16}} \)
\( = \frac{\frac{7}{16}}{\frac{25}{16}} \)
\( = \frac{7}{16} \times \frac{16}{25} \)
\( = \frac{7}{25} \)
Now, let's evaluate the Right Hand Side (R.H.S.):
\( \text{R.H.S.} = \cos^2 A - \sin^2 A \)
\( = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 \)
\( = \frac{16}{25} - \frac{9}{25} \)
\( = \frac{16 - 9}{25} \)
\( = \frac{7}{25} \)
Since L.H.S. = R.H.S., the given equation is true.
In simple words: First, use the given cotangent value to find the sides of a right triangle and then calculate sine, cosine, and tangent. Then, calculate both sides of the main equation separately. If both sides give the same answer, the equation is correct.

🎯 Exam Tip: For problems checking trigonometric identities, always calculate the Left Hand Side (LHS) and Right Hand Side (RHS) independently. Drawing a right-angled triangle helps visualize the ratios and avoid errors in calculations.

Question 3. Evaluate \( \frac{\sin 30^{\circ} + \tan 45^{\circ} - \text{cosec } 60^{\circ}}{\sec 30^{\circ} + \cos 60^{\circ} + \cot 45^{\circ}} \)
Answer:
First, we recall the standard trigonometric values:
\( \sin 30^{\circ} = \frac{1}{2} \)
\( \tan 45^{\circ} = 1 \)
\( \text{cosec } 60^{\circ} = \frac{2}{\sqrt{3}} \)
\( \sec 30^{\circ} = \frac{2}{\sqrt{3}} \)
\( \cos 60^{\circ} = \frac{1}{2} \)
\( \cot 45^{\circ} = 1 \)
Now, substitute these values into the given expression:
\( \frac{\sin 30^{\circ} + \tan 45^{\circ} - \text{cosec } 60^{\circ}}{\sec 30^{\circ} + \cos 60^{\circ} + \cot 45^{\circ}} \)
\( = \frac{\frac{1}{2} + 1 - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1} \)
Combine the terms in the numerator and denominator:
Numerator: \( \frac{1}{2} + 1 - \frac{2}{\sqrt{3}} = \frac{1+2}{2} - \frac{2}{\sqrt{3}} = \frac{3}{2} - \frac{2}{\sqrt{3}} = \frac{3\sqrt{3} - 4}{2\sqrt{3}} \)
Denominator: \( \frac{2}{\sqrt{3}} + \frac{1}{2} + 1 = \frac{2}{\sqrt{3}} + \frac{1+2}{2} = \frac{2}{\sqrt{3}} + \frac{3}{2} = \frac{4 + 3\sqrt{3}}{2\sqrt{3}} \)
Now, substitute these back into the main fraction:
\( = \frac{\frac{3\sqrt{3} - 4}{2\sqrt{3}}}{\frac{4 + 3\sqrt{3}}{2\sqrt{3}}} \)
Since the denominators \( 2\sqrt{3} \) cancel out:
\( = \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4} \)
To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is \( 3\sqrt{3} - 4 \):
\( = \frac{(3\sqrt{3} - 4)(3\sqrt{3} - 4)}{(3\sqrt{3} + 4)(3\sqrt{3} - 4)} \)
Using the identities \( (a-b)^2 = a^2 - 2ab + b^2 \) and \( (a+b)(a-b) = a^2 - b^2 \):
Numerator: \( (3\sqrt{3} - 4)^2 = (3\sqrt{3})^2 - 2(3\sqrt{3})(4) + 4^2 = 27 - 24\sqrt{3} + 16 = 43 - 24\sqrt{3} \)
Denominator: \( (3\sqrt{3})^2 - 4^2 = 27 - 16 = 11 \)
So the expression becomes:
\( = \frac{43 - 24\sqrt{3}}{11} \)
In simple words: First, write down the known values for each trigonometric function at the given angles. Then, put these values into the main fraction. Simplify the top and bottom parts separately. Finally, simplify the whole fraction by removing common terms and rationalizing the denominator.

🎯 Exam Tip: Memorize the standard trigonometric values for common angles like \( 30^{\circ}, 45^{\circ}, 60^{\circ} \). Be careful with algebraic manipulation, especially when dealing with square roots and rationalizing the denominator.

Question 4. Find A if \( \sin 20^{\circ} \tan A \sec 70^{\circ} = \sqrt{3} \)
Answer:
Given the equation: \( \sin 20^{\circ} \tan A \sec 70^{\circ} = \sqrt{3} \)
We know the complementary angle identities: \( \sin \theta = \cos (90^{\circ} - \theta) \) and \( \sec \theta = \text{cosec } (90^{\circ} - \theta) \).
Also, \( \sec \theta = \frac{1}{\cos \theta} \).
Let's rewrite \( \sin 20^{\circ} \) using a complementary angle:
\( \sin 20^{\circ} = \sin (90^{\circ} - 70^{\circ}) = \cos 70^{\circ} \)
Substitute this into the equation:
\( \cos 70^{\circ} \tan A \sec 70^{\circ} = \sqrt{3} \)
Now, rewrite \( \sec 70^{\circ} \) as \( \frac{1}{\cos 70^{\circ}} \):
\( \cos 70^{\circ} \tan A \left(\frac{1}{\cos 70^{\circ}}\right) = \sqrt{3} \)
The \( \cos 70^{\circ} \) terms cancel each other out:
\( \tan A = \sqrt{3} \)
We know that \( \tan 60^{\circ} = \sqrt{3} \).
Therefore, \( A = 60^{\circ} \).
In simple words: Use the rule that sine of an angle is cosine of its complementary angle. Then use the rule that secant is 1 over cosine. This will help you simplify the equation until only tangent of A is left. From there, find the angle A.

🎯 Exam Tip: When angles like \( 20^{\circ} \) and \( 70^{\circ} \) appear together (and add up to \( 90^{\circ} \)), always look for opportunities to use complementary angle identities to simplify the expression. This is a common trick in trigonometry problems.

Question 5. Find the area of the right triangle with hypotenuse 8 cm and one of the acute angles is \( 57^{\circ} \)
Answer:
Let the right triangle be ABC, with the right angle at B.
The hypotenuse AC = 8 cm.
One acute angle, let's say \( \angle C = 57^{\circ} \).
We need to find the other two sides, AB and BC, to calculate the area.
Using sine function for angle C:
\( \sin C = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{\text{AB}}{\text{AC}} \)
\( \sin 57^{\circ} = \frac{\text{AB}}{8} \)
We know \( \sin 57^{\circ} \approx 0.8387 \).
\( 0.8387 = \frac{\text{AB}}{8} \)
\( \implies \text{AB} = 0.8387 \times 8 \)
\( \text{AB} \approx 6.7096 \text{ cm} \)
Rounded to two decimal places, AB \( \approx 6.71 \text{ cm} \).
Using cosine function for angle C:
\( \cos C = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{\text{BC}}{\text{AC}} \)
\( \cos 57^{\circ} = \frac{\text{BC}}{8} \)
We know \( \cos 57^{\circ} \approx 0.5446 \).
\( 0.5446 = \frac{\text{BC}}{8} \)
\( \implies \text{BC} = 0.5446 \times 8 \)
\( \text{BC} \approx 4.3568 \text{ cm} \)
Rounded to two decimal places, BC \( \approx 4.36 \text{ cm} \).
Now, calculate the area of the right triangle:
Area \( = \frac{1}{2} \times \text{base} \times \text{height} \)
Area \( = \frac{1}{2} \times \text{BC} \times \text{AB} \)
Area \( = \frac{1}{2} \times 4.36 \text{ cm} \times 6.71 \text{ cm} \)
Area \( = \frac{1}{2} \times 29.2556 \text{ cm}^2 \)
Area \( \approx 14.6278 \text{ cm}^2 \)
Rounded to two decimal places, Area \( \approx 14.63 \text{ cm}^2 \).
In simple words: First, draw the triangle. Use sine and cosine with the given angle and hypotenuse to find the lengths of the two shorter sides of the triangle. Once you have the base and height, you can use the formula \( \frac{1}{2} \times \text{base} \times \text{height} \) to find the area.

🎯 Exam Tip: Always remember that the area of a right-angled triangle is half the product of its perpendicular sides. If sides are unknown, use sine and cosine functions with the given angle and hypotenuse to find them.