CBSE Class 11 Mathematics Trigonometric Functions Worksheet Set 03

Trigonometric Functions MCQ Questions with Answers Class 11 Mathematics

Question. Find the value of \( \cos(18^\circ), \cos(36^\circ), \sin(36^\circ), \sin(54^\circ) \)?
Answer: We know, \( \cos \theta = \sqrt{1 - \sin^2 \theta} \), put \( \theta = 18^\circ \)
\( \Rightarrow \cos(18^\circ) = \sqrt{1 - \sin^2(18^\circ)} \)
\( \Rightarrow \cos(18^\circ) = \sqrt{1 - \left(\frac{\sqrt{5}-1}{4}\right)^2} \)............ \( \left\{\sin(18^\circ) = \frac{\sqrt{5}-1}{4}\right\} \)
\( = \sqrt{1 - \left(\frac{5+1-2\sqrt{5}}{16}\right)} \)
\( = \sqrt{\frac{16-6+2\sqrt{5}}{16}} \)
\( = \cos(18^\circ) = \frac{\sqrt{10+2\sqrt{5}}}{4} \) ans.
Now, \( \cos(36^\circ) \)
\( = \cos 2\theta = 1 - 2\sin^2 \theta \), put \( \theta = 18^\circ \)
\( \Rightarrow \cos(36^\circ) = 1 - 2 \sin^2(18^\circ) \)
\( \Rightarrow \cos(36^\circ) = 1 - 2 \left(\frac{\sqrt{5}-1}{4}\right)^2 \)............ \( \left\{\sin(18^\circ) = \frac{\sqrt{5}-1}{4}\right\} \)
\( \Rightarrow 1- 2\left(\frac{5+1-2\sqrt{5}}{16}\right) \)
\( \Rightarrow \frac{16-12+4\sqrt{5}}{16} \)
\( = \frac{4+4\sqrt{5}}{16} \)
\( \cos(36^\circ) = \frac{\sqrt{5}+1}{4} \) ans.
Now, \( \sin(36^\circ) \)
We have, \( \sin \theta = \sqrt{1 - \cos^2 \theta} \), put \( \theta = 36^\circ \)
\( \Rightarrow \sin(36^\circ) = \sqrt{1 - \cos^2(36^\circ)} \)
\( \Rightarrow \sin(36^\circ) = \sqrt{1 - \left(\frac{\sqrt{5}+1}{4}\right)^2} \)............ \( \left\{\cos(36^\circ) = \frac{\sqrt{5}+1}{4}\right\} \)
\( \Rightarrow \sqrt{1 - \left(\frac{5+1+2\sqrt{5}}{16}\right)} \)
\( = \frac{\sqrt{16-6-2\sqrt{5}}}{4} \)
\( \sin(36^\circ) = \frac{\sqrt{10-2\sqrt{5}}}{4} \) ans.
Now, \( \sin(54^\circ) = \sin(90^\circ - 36^\circ) \)
\( = \cos(36^\circ) = \frac{\sqrt{5}+1}{4} \) ans.

Question. Find the value of \( \tan(9^\circ) - \tan(27^\circ) - \tan(63^\circ) + \tan(81^\circ) \)?
Answer: We have, \( \tan(9^\circ) - \tan(27^\circ) - \tan(63^\circ) + \tan(81^\circ) \)
\( = \tan(9^\circ) - \tan(27^\circ) - \tan(90^\circ - 27^\circ) + \tan(90^\circ - 9^\circ) \)
\( = \tan(9^\circ) - \tan(27^\circ) - \cot(27^\circ) + \cot(9^\circ) \)
\( = (\tan 9^\circ + \cot 9^\circ) - (\tan 27^\circ + \cot 27^\circ) \)
\( = \left( \frac{\sin 9^\circ}{\cos 9^\circ} + \frac{\cos 9^\circ}{\sin 9^\circ} \right) - \left( \frac{\sin 27^\circ}{\cos 27^\circ} + \frac{\cos 27^\circ}{\sin 27^\circ} \right) \)
\( = \left( \frac{\sin^2 9^\circ + \cos^2 9^\circ}{\sin 9^\circ \cos 9^\circ} \right) - \left( \frac{\sin^2 27^\circ + \cos^2 27^\circ}{\sin 27^\circ \cos 27^\circ} \right) \)
\( = \frac{1}{\sin 9^\circ \cos 9^\circ} - \frac{1}{\sin 27^\circ \cos 27^\circ} \)
\( = \frac{2}{2\sin 9^\circ \cos 9^\circ} - \frac{2}{2\sin 27^\circ \cos 27^\circ} \) {multiply and divide by 2}
\( = \frac{2}{\sin 18^\circ} - \frac{2}{\sin 54^\circ} \)
\( = \frac{2}{\sin 18^\circ} - \frac{2}{\sin(90^\circ - 36^\circ)} \)
\( = \frac{2}{\sin 18^\circ} - \frac{2}{\cos 36^\circ} \)
\( = \frac{2}{\frac{\sqrt{5}-1}{4}} - \frac{2}{\frac{\sqrt{5}+1}{4}} \)
\( = \frac{8}{\sqrt{5}-1} - \frac{8}{\sqrt{5}+1} \)
\( = \frac{8\sqrt{5}+8 - 8\sqrt{5}+8}{5-1} \)
\( = \frac{16}{4} = 4 \) ans.

Question. Simplify \( \frac{\cos x}{1+\sin x} \).
Answer: We have, \( \frac{\cos x}{1+\sin x} \)
\( = \frac{\sin\left(\frac{\pi}{2}-x\right)}{1+\cos\left(\frac{\pi}{2}-x\right)} \)
\( = \frac{2\sin\left(\frac{\pi}{4}-\frac{x}{2}\right) \cos\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)} \)............ \( \left\{\sin(\theta) = 2 \sin\frac{\theta}{2} \cdot \cos\frac{\theta}{2}\right\} \left\{1 + \cos\theta = 2\cos^2\frac{\theta}{2}\right\} \)
\( = \tan\left(\frac{\pi}{4}-\frac{x}{2}\right) \)

Question. If \( \tan x = \frac{b}{a} \), find the value of \( \sqrt{\frac{a-b}{a+b}} + \sqrt{\frac{a+b}{a-b}} \).
Answer: We have, \( \sqrt{\frac{a-b}{a+b}} + \sqrt{\frac{a+b}{a-b}} \)
\( = \sqrt{\frac{a-b}{a+b}} \times \sqrt{\frac{a-b}{a-b}} + \sqrt{\frac{a+b}{a-b}} \times \sqrt{\frac{a+b}{a+b}} \)
\( = \frac{a-b}{\sqrt{a^2-b^2}} + \frac{a+b}{\sqrt{a^2-b^2}} \)
\( = \frac{a-b+a+b}{\sqrt{a^2-b^2}} \)
\( = \frac{2a}{\sqrt{a^2-b^2}} \)
divide N & D by \( a \)
\( = \frac{\frac{2a}{a}}{\frac{\sqrt{a^2-b^2}}{a}} = \frac{2}{\sqrt{\frac{a^2}{a^2} - \frac{b^2}{a^2}}} \)
\( = \frac{2}{\sqrt{1 - \frac{b^2}{a^2}}} \)
Put \( \tan x = \frac{b}{a} \), given
\( = \frac{2}{\sqrt{1-\tan^2 x}} = \frac{2}{\sqrt{1 - \frac{\sin^2 x}{\cos^2 x}}} \)
\( = \frac{2\cos x}{\sqrt{\cos^2 x - \sin^2 x}} \)
\( = \frac{2\cos x}{\sqrt{\cos(2x)}} \) ans. \( \{\cos(2\theta) = \cos^2 \theta - \sin^2 \theta\} \)

Question. Simplify \( \frac{1-\cos x}{\sin x} \).
Answer: We have, \( \frac{1-\cos x}{\sin x} \)
\( = \frac{2\sin^2\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right) \cdot \cos\left(\frac{x}{2}\right)} \)
\( = \tan\left(\frac{x}{2}\right) \)

Question. Show that, \( \sqrt{3} \csc(20^\circ) - \sec(20^\circ) = 4 \)?
Answer: L.H.S. \( \sqrt{3} \csc(20^\circ) - \sec(20^\circ) \)
\( = \frac{\sqrt{3}}{\sin(20^\circ)} - \frac{1}{\cos(20^\circ)} \)
\( = \frac{\sqrt{3} \cos(20^\circ) - \sin(20^\circ)}{\sin(20^\circ)\cos(20^\circ)} \)
\( = \frac{2\left\{ \frac{\sqrt{3}}{2}\cos(20^\circ) - \frac{1}{2}\sin(20^\circ) \right\}}{\frac{1}{2}(2\sin(20^\circ)\cos(20^\circ))} \)
\( = \frac{2\{\sin(60^\circ)\cos(20^\circ) - \cos(60^\circ)\sin(20^\circ)\}}{\frac{1}{2}\sin(40^\circ)} \) \( \{\sin(2\theta) = 2\sin\theta\cos\theta\} \)
\( = \frac{4\sin(60^\circ-20^\circ)}{\sin(40^\circ)} = \frac{4\sin(40^\circ)}{\sin(40^\circ)} = 4 \) R.H.S. ans.

Question. Simplify \( \frac{1-\sin x}{\cos x} \).
Answer: We have, \( \frac{1-\sin x}{\cos x} \)
\( = \frac{1-\cos\left(\frac{\pi}{2}-x\right)}{\sin\left(\frac{\pi}{2}-x\right)} \)
\( = \frac{2\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2\sin\left(\frac{\pi}{4}-\frac{x}{2}\right)\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)} \)
\( = \tan\left(\frac{\pi}{4}-\frac{x}{2}\right) \)

Question. Show that, \( \frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \frac{1+\sin A}{\cos A} \)?
Answer: L.H.S. \( \frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} \)
\( = \frac{\frac{\sin A}{\cos A} + \frac{1}{\cos A} - 1}{\frac{\sin A}{\cos A} - \frac{1}{\cos A} + 1} \)
\( = \frac{\sin A + 1 - \cos A}{\sin A - 1 + \cos A} \)
\( = \frac{\sin A + (1 - \cos A)}{\sin A - (1 - \cos A)} \)
\( = \frac{\sin A + 2\sin^2\frac{A}{2}}{\sin A - 2\sin^2\frac{A}{2}} \) \( \left\{1 - \cos\theta = 2\sin^2\frac{\theta}{2}\right\} \)
\( = \frac{2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right) + 2\sin^2\left(\frac{A}{2}\right)}{2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right) - 2\sin^2\left(\frac{A}{2}\right)} \) \( \left\{\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right\} \)
\( = \frac{2\sin\left(\frac{A}{2}\right)\left[\cos\left(\frac{A}{2}\right) + \sin\left(\frac{A}{2}\right)\right]}{2\sin\left(\frac{A}{2}\right)\left[\cos\left(\frac{A}{2}\right) - \sin\left(\frac{A}{2}\right)\right]} \)
Rationalize as we have to make the angle to \( A \) from \( \frac{A}{2} \)
\( = \frac{\left[\cos\left(\frac{A}{2}\right) + \sin\left(\frac{A}{2}\right)\right]\left[\cos\left(\frac{A}{2}\right) + \sin\left(\frac{A}{2}\right)\right]}{\left[\cos\left(\frac{A}{2}\right) - \sin\left(\frac{A}{2}\right)\right]\left[\cos\left(\frac{A}{2}\right) + \sin\left(\frac{A}{2}\right)\right]} \)
\( = \frac{\cos^2\left(\frac{A}{2}\right) + \sin^2\left(\frac{A}{2}\right) + 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)}{\cos^2\left(\frac{A}{2}\right) - \sin^2\left(\frac{A}{2}\right)} \)
\( = \frac{1 + \sin A}{\cos A} \) R.H.S (proved) \( \left\{1 - \cos\theta = 2\sin^2\frac{\theta}{2}, \cos(2\theta) = \cos^2\theta - \sin^2\theta\right\} \)

Question. Show that \( \sin(4A) = 4\sin A \cos^3 A - 4\cos A \sin^3 A \)?
Answer: HINT: \( \sin(4A) = 2\sin(2A)\cos(2A) \)

Question. Show that, \( \cos^2(A-B) + \cos^2 B - 2\cos(A-B)\cos A \cos B = \sin^2 A \)?
Answer: L.H.S. \( \cos^2(A-B) + \cos^2 B - 2\cos(A-B)\cos A \cos B \)
\( = \cos(A-B) [\cos(A-B) - 2\cos A \cos B] + \cos^2 B \)
\( = \cos(A-B) [\cos A \cos B + \sin A \sin B - 2\cos A \cos B] + \cos^2 B \)
\( = \cos(A-B) [-\cos A \cos B + \sin A \sin B] + \cos^2 B \) ........ (- common for making a formula)
\( = -\cos(A-B) [\cos A \cos B - \sin A \sin B] + \cos^2 B \)
\( = -\cos(A-B)\cos(A+B) + \cos^2 B \)
\( = -[\cos^2 A - \sin^2 B] + \cos^2 B \) ............. \( [\cos(A+B)\cos(A-B) = \cos^2 A - \sin^2 B] \)
\( = -\cos^2 A + (\sin^2 B + \cos^2 B) \)
\( = -\cos^2 A + 1 = 1 - \cos^2 A \) ............. \( \left\{\cos^2\theta + \sin^2\theta = 1\right\} \)
\( = \sin^2 A = \) R.H.S. (proved)

Question. Show that, \( \cos A \cos(2A) \cos(2^2 A) \cos(2^3 A) \dots \cos(2^{n-1} A) = \frac{\sin(2^n A)}{2^n \sin A} \)?
Answer: Taking, L.H.S. \( \cos A \cos(2A) \cos(2^2 A) \cos(2^3 A) \dots \cos(2^{n-1} A) \)
Multiply & divide by \( 2\sin A \)
\( = \frac{1}{2\sin A} \left[(2\sin A \cos A) \cos(2A) \cos(2^2 A) \cos(2^3 A) \dots \cos(2^{n-1} A)\right] \)
\( = \frac{1}{2\sin A} \left[\sin(2A) \cos(2A) \cos(2^2 A) \cos(2^3 A) \dots \cos(2^{n-1} A)\right] \)
Multiply & divide by 2
\( = \frac{1}{2^2 \sin A} \left[2\sin(2A) \cos(2A) \cos(2^2 A) \cos(2^3 A) \dots \cos(2^{n-1} A)\right] \)
\( = \frac{1}{2^2 \sin A} \left[(\sin(4A) \cos(4A)) \cos(2^3 A) \dots \cos(2^{n-1} A)\right] \)
\( = \frac{1}{2^3 \sin A} \left[(2\sin(4A) \cos(4A)) \cos(2^3 A) \dots \cos(2^{n-1} A)\right] \)
\( = \frac{1}{2^3 \sin A} \left[\sin(2^3 A) \cos(2^3 A) \dots \cos(2^{n-1} A)\right] \)
Once the process goes on.......
\( = \frac{1}{2^{n-1} \sin A} \left[\sin(2^{n-1} A) \cos(2^{n-1} A)\right] \)
Multiply & divide by 2
\( = \frac{1}{2^n \sin A} \left[2\sin(2^{n-1} A) \cos(2^{n-1} A)\right] \)
\( = \frac{1}{2^n \sin A} [\sin(2 \cdot 2^{n-1} A)] \) ...................... \( \{\sin(2\theta) = 2\sin\theta\cos\theta\} \)
\( = \frac{1}{2^n \sin A} \cdot \sin(2^n A) \). R.H.S. (proved)

Question. Show that, \( \cos\left(\frac{\pi}{7}\right) \cos\left(\frac{2\pi}{7}\right) \cos\left(\frac{4\pi}{7}\right) = -\frac{1}{8} \)?
Answer: L.H.S. \( \cos\left(\frac{\pi}{7}\right) \cos\left(\frac{2\pi}{7}\right) \cos\left(\frac{4\pi}{7}\right) \)
Multiply & divide by \( 2\sin\left(\frac{\pi}{7}\right) \)
\( = \frac{1}{2\sin\left(\frac{\pi}{7}\right)} \left[ \left(2\sin\left(\frac{\pi}{7}\right) \cos\left(\frac{\pi}{7}\right)\right) \cos\left(\frac{2\pi}{7}\right) \cos\left(\frac{4\pi}{7}\right) \right] \)
\( = \frac{1}{2\sin\left(\frac{\pi}{7}\right)} \left[\sin\left(\frac{2\pi}{7}\right) \cos\left(\frac{2\pi}{7}\right) \cos\left(\frac{4\pi}{7}\right)\right] \)
\( = \frac{1}{2^2 \sin\left(\frac{\pi}{7}\right)} \left[ \left(2\sin\frac{2\pi}{7} \cos\frac{2\pi}{7}\right) \cos\left(\frac{4\pi}{7}\right) \right] \)
\( = \frac{1}{2^2 \sin\left(\frac{\pi}{7}\right)} \left[\sin\left(\frac{4\pi}{7}\right) \cos\left(\frac{4\pi}{7}\right)\right] \)
\( = \frac{1}{2^3 \sin\left(\frac{\pi}{7}\right)} \left[2\sin\left(\frac{4\pi}{7}\right) \cos\left(\frac{4\pi}{7}\right)\right] \)
\( = \frac{1}{2^3 \sin\left(\frac{\pi}{7}\right)} \cdot \sin\left(\frac{8\pi}{7}\right) \)
\( = \frac{1}{8\sin\left(\frac{\pi}{7}\right)} \cdot \sin\left(\pi + \frac{\pi}{7}\right) \)
\( = \frac{1}{8\sin\left(\frac{\pi}{7}\right)} \cdot \left(-\sin\left(\frac{\pi}{7}\right)\right) \)
\( = -\frac{1}{8} \) R.H.S. (proved)

Question. Show that, \( \cos\left(\frac{2\pi}{15}\right) \cos\left(\frac{4\pi}{15}\right) \cos\left(\frac{8\pi}{15}\right) \cos\left(\frac{14\pi}{15}\right) = \frac{1}{16} \)?
Answer: L.H.S. \( \cos\left(\frac{2\pi}{15}\right) \cos\left(\frac{4\pi}{15}\right) \cos\left(\frac{8\pi}{15}\right) \cos\left(\frac{14\pi}{15}\right) \)
\( = \cos\left(\frac{2\pi}{15}\right) \cos\left(\frac{4\pi}{15}\right) \cos\left(\frac{8\pi}{15}\right) \cos\left(\pi - \frac{\pi}{15}\right) \)
\( = \cos\left(\frac{2\pi}{15}\right) \cos\left(\frac{4\pi}{15}\right) \cos\left(\frac{8\pi}{15}\right) \left(-\cos\frac{\pi}{15}\right) \)
\( = -\cos\left(\frac{\pi}{15}\right) \cos\left(\frac{2\pi}{15}\right) \cos\left(\frac{4\pi}{15}\right) \cos\left(\frac{8\pi}{15}\right) \)
Multiply & divide by \( 2\sin\left(\frac{\pi}{15}\right) \)
\( = \frac{-1}{2\sin\left(\frac{\pi}{15}\right)} \left[ 2\sin\left(\frac{\pi}{15}\right) \cos\left(\frac{\pi}{15}\right) \cos\left(\frac{2\pi}{15}\right) \cos\left(\frac{4\pi}{15}\right) \cos\left(\frac{8\pi}{15}\right) \right] \)
\( = \frac{-1}{2\sin\left(\frac{\pi}{15}\right)} \left[ 2\sin\left(\frac{2\pi}{15}\right) \cos\left(\frac{2\pi}{15}\right) \cos\left(\frac{4\pi}{15}\right) \cos\left(\frac{8\pi}{15}\right) \right] \)
Proceed as Q.11
\( = \frac{-1}{2^4 \sin\left(\frac{\pi}{15}\right)} \cdot \sin\left(\frac{16\pi}{15}\right) \)
\( = \frac{-1}{2^4 \sin\left(\frac{\pi}{15}\right)} \cdot \sin\left(\pi + \frac{\pi}{15}\right) \)
\( = \frac{-1}{16\sin\left(\frac{\pi}{15}\right)} \cdot \left(-\sin\left(\frac{\pi}{15}\right)\right) \)
\( = \frac{1}{16} \) R.H.S. (proved) ans.

Question. Show that, \( \cot A + \cot(60^\circ + A) - \cot(60^\circ - A) = 3\cot(3A) \)?
Answer: L.H.S. \( \cot A + \cot(60^\circ + A) - \cot(60^\circ - A) \)
\( = \frac{1}{\tan A} + \frac{1}{\tan(60^\circ+A)} - \frac{1}{\tan(60^\circ-A)} \)
\( = \frac{1}{\tan A} + \frac{1-\sqrt{3}\tan A}{\sqrt{3}+\tan A} - \frac{1+\sqrt{3}\tan A}{\sqrt{3}-\tan A} \) .................... \( [\tan(A+B) \cdot \tan(A-B) \text{ formula}] \)
\( = \frac{1}{\tan A} + \left[ \frac{(1-\sqrt{3}\tan A)(\sqrt{3}-\tan A) - (1+\sqrt{3}\tan A)(\sqrt{3}+\tan A)}{(\sqrt{3}+\tan A)(\sqrt{3}-\tan A)} \right] \)
\( = \frac{1}{\tan A} + \left[ \frac{\sqrt{3} - \tan A - 3\tan A + \sqrt{3}\tan^2 A - \sqrt{3} - \tan A - 3\tan A - \sqrt{3}\tan^2 A}{3-\tan^2 A} \right] \)
\( = \frac{1}{\tan A} + \frac{-8\tan A}{3-\tan^2 A} \)
\( = \frac{3-\tan^2 A - 8\tan^2 A}{\tan A(3-\tan^2 A)} \)
\( = \frac{3-9\tan^2 A}{3\tan A - \tan^3 A} \)
\( = \frac{3(1-3\tan^2 A)}{3\tan A - \tan^3 A} \)
\( = \frac{3}{\tan(3A)} \) ..................... \( \left\{\tan(3\theta) = \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\right\} \)
\( = 3\cot(3A) = \) R.H.S. (proved) ans.

Question. Show that, \( 2\sin^2 \beta + 4\cos(\alpha+\beta)\sin\alpha\sin\beta + \cos(2\alpha+2\beta) = \cos(2\alpha) \)?
Answer: L.H.S. \( 2\sin^2 \beta + 4\cos(\alpha+\beta)\sin\alpha\sin\beta + \cos(2\alpha+2\beta) \)
\( = 2\sin^2 \beta + 4(\cos\alpha\cos\beta - \sin\alpha\sin\beta)\sin\alpha\sin\beta + (\cos(2\alpha)\cos(2\beta) - \sin(2\alpha)\sin(2\beta)) \)
\( = 2\sin^2 \beta + 4\cos\alpha\cos\beta\sin\alpha\sin\beta - 4\sin^2\alpha\sin^2\beta + \cos(2\alpha)\cos(2\beta) - \sin(2\alpha)\sin(2\beta) \)
\( = 2\sin^2\beta + (2\sin\alpha\cos\alpha)(2\sin\beta\cos\beta) - 4\sin^2\alpha\sin^2\beta + \cos(2\alpha)\cos(2\beta) - \sin(2\alpha)\sin(2\beta) \)
\( = 2\sin^2\beta + \sin(2\alpha)\sin(2\beta) - 4\sin^2\alpha\sin^2\beta + \cos(2\alpha)\cos(2\beta) - \sin(2\alpha)\sin(2\beta) \)
\( = (1 - \cos 2\beta) - (1-\cos 2\alpha)(1-\cos 2\beta) + \cos(2\alpha)\cos(2\beta) \) ............. \( \left\{\sin^2\theta = \frac{1-\cos(2\theta)}{2}\right\} \)
\( = 1 - \cos 2\beta - (1 - \cos 2\beta - \cos 2\alpha + \cos 2\alpha\cos 2\beta) + \cos(2\alpha)\cos(2\beta) \)
\( = 1 - \cos 2\beta - 1 + \cos 2\beta + \cos 2\alpha - \cos 2\alpha\cos 2\beta + \cos(2\alpha)\cos(2\beta) \)
\( = \cos(2\alpha) = \) R.H.S. (proved) ans.

Question. If \( x \cos\theta = y \cos\left(\theta + \frac{2\pi}{3}\right) = z \cos\left(\theta + \frac{4\pi}{3}\right) \), then find the value of \( xy + yz + zx \)?
Answer: We can write, \( xy + yz + zx = xyz \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \)
Let \( x \cos\theta = y \cos\left(\theta + \frac{2\pi}{3}\right) = z \cos\left(\theta + \frac{4\pi}{3}\right) = k \text{ (say)} \)
Then, \( \frac{1}{x} = \frac{\cos\theta}{k}, \frac{1}{y} = \frac{\cos\left(\theta + \frac{2\pi}{3}\right)}{k}, \frac{1}{z} = \frac{\cos\left(\theta + \frac{4\pi}{3}\right)}{k} \)
Now, \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{k} \left[\cos\theta + \cos\left(\theta + \frac{2\pi}{3}\right) + \cos\left(\theta + \frac{4\pi}{3}\right)\right] \)
\( = \frac{1}{k} \left[\cos\theta + \cos\theta\cos\frac{2\pi}{3} - \sin\theta\sin\frac{2\pi}{3} + \cos\theta\cos\frac{4\pi}{3} - \sin\theta\sin\left(\frac{4\pi}{3}\right)\right] \)
\( = \frac{1}{k} \left[\cos\theta + \cos\theta\cos\left(\pi - \frac{\pi}{3}\right) - \sin\theta\sin\left(\pi - \frac{\pi}{3}\right) + \cos\theta\cos\left(\pi + \frac{\pi}{3}\right) - \sin\theta\sin\left(\pi + \frac{\pi}{3}\right)\right] \)
\( = \frac{1}{k} \left[\cos\theta + \cos\theta \left(-\frac{1}{2}\right) - \sin\theta \left(\frac{\sqrt{3}}{2}\right) + \cos\theta \left(-\frac{1}{2}\right) - \sin\theta \left(-\frac{\sqrt{3}}{2}\right)\right] \)
\( = \frac{1}{k} \left[\cos\theta - \frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta - \frac{1}{2}\cos\theta + \frac{\sqrt{3}}{2}\sin\theta\right] \)
\( = \frac{1}{k} [\cos\theta - \cos\theta] = 0 \)
\( \therefore \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \)
Since, \( xy + yz + zx = xyz \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \)
\( = xyz \times (0) = 0 = \) R.H.S. ans.

Question. Find the value of expansion \( 3\left[\sin^4\left(\frac{3\pi}{2} - \alpha\right) + \sin^4(3\pi + \alpha)\right] - 2\left[\sin^6\left(\frac{\pi}{2} + \alpha\right) + \sin^6(5\pi + \alpha)\right] \)?
Answer: \( 3\left[\sin^4\left(\frac{3\pi}{2} - \alpha\right) + \sin^4(3\pi + \alpha)\right] - 2\left[\sin^6\left(\frac{\pi}{2} + \alpha\right) + \sin^6(5\pi + \alpha)\right] \)
\( = 3\left[(\cos^2 \alpha + \sin^2 \alpha)^2 - 2\cos^2 \alpha \sin^2 \alpha\right] - 2\left[(\cos^2\alpha + \sin^2\alpha)(\cos^4\alpha + \sin^4\alpha) - \cos^2\alpha\sin^2\alpha\right] \)
\( = 3[1 - 2\cos^2\alpha\sin^2\alpha] - 2\left[(1)(\cos^2\alpha+\sin^2\alpha)^2 - 2\cos^2\alpha\sin^2\alpha - \cos^2\alpha\sin^2\alpha\right] \)
\( = 3 - 6\cos^2\alpha\sin^2\alpha - 2[1 - 3\cos^2\alpha\sin^2\alpha] \)
\( = 3 - 6\cos^2\alpha\sin^2\alpha - 2 + 6\cos^2\alpha\sin^2\alpha \)
\( = 3 - 2 = 1 \) ans.

Question. If \( \cos\alpha + \cos\beta = 0 = \sin\alpha + \sin\beta \), then show that \( \cos(2\alpha) + \cos(2\beta) = -2\cos(\alpha+\beta) \)?
Answer: We have, \( \cos\alpha + \cos\beta = \sin\alpha + \sin\beta = 0 \)
Then also we have, \( (\cos\alpha + \cos\beta)^2 - (\sin\alpha + \sin\beta)^2 = 0 \)
\( \Rightarrow \cos^2\alpha + \cos^2\beta + 2\cos\alpha\cos\beta - (\sin^2\alpha + \sin^2\beta + 2\sin\alpha\sin\beta) = 0 \)
\( \Rightarrow (\cos^2\alpha - \sin^2\alpha) + (\cos^2\beta - \sin^2\beta) + 2(\cos\alpha\cos\beta - \sin\alpha\sin\beta) = 0 \)
\( \Rightarrow \cos(2\alpha) + \cos(2\beta) + 2\cos(\alpha+\beta) = 0 \)............. \( \{\cos(2\theta) = \cos^2\theta - \sin^2\theta\} \)
\( \Rightarrow \cos(2\alpha) + \cos(2\beta) = -2\cos(\alpha+\beta) \) (proved)

Question. If \( \tan\frac{\theta}{2} = \sqrt{\frac{1-e}{1+e}}\tan\left(\frac{\phi}{2}\right) \), show that \( \cos\phi = \frac{\cos\theta-e}{1-e\cos\theta} \)?
Answer: We have, \( \tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{1-e}{1+e}}\tan\left(\frac{\phi}{2}\right) \)
Squaring, \( \tan^2\left(\frac{\theta}{2}\right) = \left(\frac{1-e}{1+e}\right)\tan^2\left(\frac{\phi}{2}\right) \)
Taking, L.H.S. \( \cos\phi \)
\( = \frac{1-\tan^2\left(\frac{\phi}{2}\right)}{1+\tan^2\left(\frac{\phi}{2}\right)} \) ..................... \( \left\{\cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta}\right\} \)
\( = \frac{1 - \left(\frac{1+e}{1-e}\right)\tan^2\left(\frac{\theta}{2}\right)}{1 + \left(\frac{1+e}{1-e}\right)\tan^2\left(\frac{\theta}{2}\right)} \) ..................... \( \left\{\text{putting value of }\tan^2\left(\frac{\phi}{2}\right)\right\} \)
\( = \frac{(1-e)-(1+e)\tan^2\left(\frac{\theta}{2}\right)}{(1-e)+(1+e)\tan^2\left(\frac{\theta}{2}\right)} \)
\( = \frac{1-e-\tan^2\frac{\theta}{2}-e\tan^2\frac{\theta}{2}}{1-e+\tan^2\frac{\theta}{2}+e\tan^2\frac{\theta}{2}} \)
\( = \frac{\left(1-\tan^2\frac{\theta}{2}\right)-e\left(1+\tan^2\frac{\theta}{2}\right)}{\left(1+\tan^2\frac{\theta}{2}\right)-e\left(1-\tan^2\frac{\theta}{2}\right)} \)
Divide N & D by \( \left(1 + \tan^2\frac{\theta}{2}\right) \)
\( = \frac{\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}} - e}{1 - e\left(\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}\right)} \)
\( = \frac{\cos\theta-e}{1-e\cos\theta} = \) R.H.S. ..................... \( \left\{\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}} = \cos\theta\right\} \) ans.