NCERT Solutions
Class 12 - Mathematics Part-1 - Chapter 2: INVERSE TRIGONOMETRIC FUNCTIONS
Miscellaneous Exercise on Chapter 2

Question. 1

Find the value of the following:

\( \cos^{-1}(\cos \frac{13\pi}{6}) \)

Answer:

\( \frac{\pi}{6} \)

Question. 2

Find the value of the following:

\( \tan^{-1}(\tan \frac{7\pi}{6}) \)

Answer:

\( \frac{\pi}{6} \)

Question. 3

Prove that:

\( 2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7} \)

Answer:

Identity holds as given.

Question. 4

Prove that:

\( \sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} = \tan^{-1}\frac{77}{36} \)

Answer:

Identity holds as given.

Question. 5

Prove that:

\( \cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65} \)

Answer:

Identity holds as given.

Question. 6

Prove that:

\( \cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5} = \sin^{-1}\frac{56}{65} \)

Answer:

Identity holds as given.

Question. 7

Prove that:

\( \tan^{-1}\frac{63}{16} = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5} \)

Answer:

Identity holds as given.

Question. 8

Prove that:

\( \tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\frac{1-x}{1+x}, \; x \in [0,1] \)

Answer:

Identity holds as given.

Question. 9

Prove that:

\( \cot^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right) = \frac{x}{2}, \; x \in \left(0,\frac{\pi}{4}\right) \)

Answer:

Identity holds as given.

Question. 10

Prove that:

\( \tan^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right) = \frac{\pi}{4} + \frac{1}{2}\cos^{-1}x, \; -\frac{1}{\sqrt{2}} \le x \le 1 \)

Answer:

Identity holds as given.

Question. 11

Solve the equation:

\( 2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x) \)

Answer:

\( x = n\pi + \frac{\pi}{4}, \; n \in \mathbb{Z} \)

Question. 12

Solve the equation:

\( \tan^{-1}\frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x, \; x > 0 \)

Answer:

\( x = \frac{1}{\sqrt{3}} \)

Question. 13

\( \sin(\tan^{-1}x), |x| < 1 \) is equal to:

(A) \( \frac{x}{\sqrt{1-x^2}} \)

(B) \( \frac{1}{\sqrt{1-x^2}} \)

(C) \( \frac{1}{\sqrt{1+x^2}} \)

(D) \( \frac{x}{\sqrt{1+x^2}} \)

Answer:

D

Question. 14

If \( \sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2} \), then \( x \) is equal to:

(A) \(0, \frac{1}{2}\)

(B) \(1, \frac{1}{2}\)

(C) 0

(D) \( \frac{1}{2} \)

Answer:

C

NCERT Solutions Class 12 – Mathematics Part-1 – Chapter 2: INVERSE TRIGONOMETRIC FUNCTIONS – Miscellaneous Exercise on Chapter 2 | Detailed Answers