CBSE Class 11 Mathematics Trigonometric Functions Worksheet Set 04

Question. Find the principal solutions of cosec 𝑥 = −2
Answer : ⇒ sin 𝑥 = −1/2
3rd quadrant, sin 𝑥 = sin (𝜋 +𝜋/6)
⇒ sin 𝑥 = sin (7𝜋/6)
4th quadrant, sin 𝑥 = sin (2𝜋 + 𝜋/6)
⇒ sin 𝑥 = sin (11𝜋/6)
⇒ 𝑥 = 7𝜋/6 and 𝑥 = 11𝜋/6 are the principal solutions

Question. Find the principal solutions of cosec 𝑥 = −2
Answer : cosec 𝑥
⇒ sin 𝑥 = −1/2
3rd quadrant, sin 𝑥 = sin (𝜋 +𝜋/6)
⇒ sin 𝑥 = sin (7𝜋/6)
4th quadrant, sin 𝑥 = sin (2𝜋 + 𝜋/6)
⇒ sin 𝑥 = sin (11𝜋/6)
⇒ 𝑥 = 7𝜋/6 and 𝑥 = 11𝜋/6 are the principal solutions

Question. Find the principal solutions of tan 𝑥 = √3
Answer : tan 𝑥 = √3
1st quadrant, tan 𝑥 = tan (𝜋/3)
⇒ 𝑥 = 𝜋/3 3rd quadrant, tan 𝑥 = tan (𝜋 +𝜋/3)
⇒ tan 𝑥 = tan (4𝜋/3)
∴ 𝑥 = 𝜋/3 and 𝑥 = 4𝜋/3 are the principal solutions

Question. Find the principal solutions of, sec 𝑥 = −1
Answer : sec 𝑥 = −1
⇒ cos 𝑥 = −1
2nd quadrant, cos 𝑥 = cos(𝜋 − 0)
⇒ 𝑥 = 𝜋
3rd quadrant, cos 𝑥 = cos(𝜋 + 0)
⇒ 𝑥 = 𝜋
∴ 𝑥 = 𝜋 is the principal solutions

Question. Solve, \( \tan\theta \tan\left(\theta + \frac{\pi}{3}\right) + \tan\left(\theta + \frac{2\pi}{3}\right) = 3 \)
Answer: We have, \( \tan\theta \tan\left(\theta + \frac{\pi}{3}\right) + \tan\left(\theta + \frac{2\pi}{3}\right) = 3 \)
\( \implies \tan\theta + \frac{\tan\theta + \tan(60^\circ)}{1 - \tan\theta\tan(60^\circ)} + \frac{\tan\theta + \tan(120^\circ)}{1 - \tan\theta\tan(120^\circ)} = 3 \)
\( \implies \tan\theta + \frac{\tan\theta + \sqrt{3}}{1 - \sqrt{3}\tan\theta} + \frac{\tan\theta - \sqrt{3}}{1 + \sqrt{3}\tan\theta} = 3 \)
\( \implies \tan\theta + \frac{8\tan\theta}{1 - 3\tan^2\theta} = 3 \) ................. (after taking L.C.M)
\( \implies \frac{\tan\theta - 3\tan^3\theta + 8\tan\theta}{1 - 3\tan^2\theta} = 3 \)
\( \implies \frac{9\tan\theta - 3\tan^3\theta}{1 - 3\tan^2\theta} = 3 \)
\( \implies \frac{3(3\tan\theta - \tan^3\theta)}{1 - 3\tan^2\theta} = 3 \)
\( \implies 3\tan(3\theta) = 3 \)
\( \implies \tan(3\theta) = 1 \)
\( \implies \tan(3\theta) = \tan\left(\frac{\pi}{4}\right) \)
\( \implies 3\theta = n\pi + \frac{\pi}{4} \)
\( \implies \theta = \frac{n\pi}{3} + \frac{\pi}{12} ; n \in \mathbb{Z} \) ans.

Question. Solve, \( 2\tan^2 x + \sec^2 x = 2 \); \( 0 \le x \le 2\pi \)
Answer: We have, \( 2\tan^2 x + \sec^2 x = 2 \); \( 0 \le x \le 2\pi \)
\( \implies 2\tan^2 x + 1 + \tan^2 x = 2 \)
\( \implies 3\tan^2 x = 1 \)
\( \implies \tan^2 x = \frac{1}{3} \)
\( \implies \tan x = \pm\frac{1}{\sqrt{3}} \)
\( \implies \tan x = \frac{1}{\sqrt{3}} \) and \( \tan x = -\frac{1}{\sqrt{3}} \)
For \( \tan x = \frac{1}{\sqrt{3}} \):
\( x = \frac{\pi}{6} \) or \( x = \pi + \frac{\pi}{6} \)
\( \implies x = \frac{\pi}{6} \) or \( x = \frac{7\pi}{6} \)
For \( \tan x = -\frac{1}{\sqrt{3}} \):
\( x = \pi - \frac{\pi}{6} \) or \( x = 2\pi - \frac{\pi}{6} \)
\( \implies x = \frac{5\pi}{6} \) or \( x = \frac{11\pi}{6} \)
\( x = \frac{\pi}{6}, x = \frac{7\pi}{6}, x = \frac{5\pi}{6}, x = \frac{11\pi}{6} \) are the possible values of the given equation where \( 0 \le x \le 2\pi \) ans.

Question. If \( \cot\theta + \tan\theta = 2\csc\theta \), solve for \( \theta \)
Answer: We have, \( \cot\theta + \tan\theta = 2\csc\theta \)
\( \implies \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} = \frac{2}{\sin\theta} \)
\( \implies \frac{\cos^2\theta + \sin^2\theta}{\sin\theta\cos\theta} = \frac{2}{\sin\theta} \)
\( \implies \frac{1}{\sin\theta\cos\theta} = \frac{2}{\sin\theta} \)
\( \implies \sin\theta = 2\sin\theta\cos\theta \)
\( \implies \sin\theta - 2\sin\theta\cos\theta = 0 \)
\( \implies \sin\theta(1 - 2\cos\theta) = 0 \)
Either \( \sin\theta = 0 \)
\( \implies \theta = n\pi \)
Or,
\( 1 - 2\cos\theta = 0 \)
\( \implies \cos\theta = \frac{1}{2} \)
\( \implies \cos\theta = \cos\frac{\pi}{3} \)
Here, \( \alpha = \frac{\pi}{3} \)
\( \implies \theta = 2n\pi \pm \frac{\pi}{3} \)
\( \therefore \theta = n\pi \) and \( \theta = 2n\pi \pm \frac{\pi}{3} ; n \in \mathbb{Z} \) ans.

Question. In any \( \Delta ABC \) show that \( a \cos\left(\frac{B-C}{2}\right) = (b+c) \sin\frac{A}{2} \)
Answer: R.H.S. \( (b + c) \sin\frac{A}{2} \)
Using sine law \( b = k\sin B \) and \( c = k\sin C \)
\( = k(\sin B + \sin C)\sin\frac{A}{2} \)
\( = k \cdot 2\sin\left(\frac{B+C}{2}\right) \cdot \cos\left(\frac{B-C}{2}\right) \cdot \sin\frac{A}{2} \) ............ {using \( \sin B + \sin C \) formula}
\( = k \cdot 2\sin\left(\frac{\pi-A}{2}\right) \cdot \cos\left(\frac{B-C}{2}\right) \cdot \sin\frac{A}{2} \) .......... {\( A+B+C=\pi \)}
\( = k \cdot 2\cos\left(\frac{A}{2}\right) \cdot \cos\left(\frac{B-C}{2}\right) \cdot \sin\frac{A}{2} \) .......... {\( \sin\left(\frac{\pi}{2} - \frac{A}{2}\right) = \cos\left(\frac{A}{2}\right) \)}
\( = k \cdot \left(2\sin\frac{A}{2} \cdot \cos\frac{A}{2}\right) \cdot \cos\left(\frac{B-C}{2}\right) \)
\( = k \cdot \sin A \cdot \cos\left(\frac{B-C}{2}\right) \) .......... {\( 2\sin\theta\cos\theta = \sin(2\theta) \)}
\( = a \cos\left(\frac{B-C}{2}\right) = \) R.H.S. (proved) ............... {since sine law \( a = k\sin A \)}

Question. In any \( \Delta ABC \) show that \( \frac{\sin(B-C)}{\sin(B+C)} = \frac{b^2-c^2}{a^2} \)
Answer: R.H.S. \( \frac{b^2 - c^2}{a^2} \)
Using sine law \( a = k\sin A, b = k\sin B \) and \( c = k\sin C \)
\( = \frac{k^2\sin^2 B - k^2\sin^2 C}{k^2\sin^2 A} \)
\( = \frac{\sin^2 B - \sin^2 C}{\sin^2 A} \)
\( = \frac{\sin(B+C) \cdot \sin(B-C)}{\sin^2 A} \) ........... {\( \sin^2 B - \sin^2 C = \sin(B+C)\sin(B-C) \)}
\( = \frac{\sin(\pi - A) \cdot \sin(B-C)}{\sin^2 A} \) ............. {\( A + B + C = \pi \)}
\( = \frac{\sin A \cdot \sin(B-C)}{\sin^2 A} \)
\( = \frac{\sin(B-C)}{\sin A} \)
\( = \frac{\sin(B-C)}{\sin(\pi - (B+C))} \)
\( = \frac{\sin(B-C)}{\sin(B+C)} = \) L.H.S. (proved)

Question. In any \( \Delta ABC \) show that \( a^3 \sin(B - C) + b^3 \sin(C - A) + c^3 \sin(A - B) = 0 \)
Answer: L.H.S. \( a^3 \sin(B - C) + b^3 \sin(C - A) + c^3 \sin(A - B) \)
Using sine law \( a = k\sin A, b = k\sin B \) and \( c = k\sin C \)
\( = k^3 [\sin^3 A \cdot \sin(B - C) + \sin^3 B \cdot \sin(C - A) + \sin^3 C \cdot \sin(A - B)] \)
\( = k^3 [\sin^2 A \cdot \sin A \cdot \sin(B - C) + \sin^2 B \cdot \sin B \cdot \sin(C - A) + \sin^2 C \cdot \sin C \cdot \sin(A - B)] \)
\( = k^3 [\sin^2 A \cdot \sin(\pi - (B+C)) \cdot \sin(B - C) + \sin^2 B \cdot \sin(\pi - (C+A)) \cdot \sin(C - A) + \sin^2 C \cdot \sin(\pi - (A+B)) \cdot \sin(A - B)] \)
\( = k^3 [\sin^2 A \cdot \sin(B+C) \cdot \sin(B - C) + \sin^2 B \cdot \sin(C+A) \cdot \sin(C - A) + \sin^2 C \cdot \sin(A+B) \cdot \sin(A - B)] \)
\( = k^3 [\sin^2 A(\sin^2 B - \sin^2 C) + \sin^2 B(\sin^2 C - \sin^2 A) + \sin^2 C(\sin^2 A - \sin^2 B)] \)
\( = k^3 [\sin^2 A \sin^2 B - \sin^2 A \sin^2 C + \sin^2 B \sin^2 C - \sin^2 A \sin^2 B + \sin^2 A \sin^2 C - \sin^2 B \sin^2 C] \)
\( = k^3 \cdot 0 = 0 = \) R.H.S. (proved) ans.