#### Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 7 Sub Question 1 Maths Textbook Solution.

Answer: $\frac{\pi}{4}$

Given: Given that  $\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}$

Hint: First we solve $2 \sin ^{-1} \frac{1}{2}$ Since $\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$

Solution: We have,

$\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}$

$=\tan ^{-1}\left\{2 \cos \left(2 \times \frac{\pi}{6}\right)\right\}$                                                            $\left[\sin ^{-1} \frac{1}{2}=\frac{\pi}{6}\right]$

\begin{aligned} &=\tan ^{-1}\left\{2 \cos \frac{\pi}{3}\right\} \\ &=\tan ^{-1}\left\{2 \times \frac{1}{2}\right\} \end{aligned}                                                                        $\left[\cos \frac{\pi}{3}=\frac{1}{2}\right]$

\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}                                                                                       $\left[\tan ^{-1}(1)=\frac{\pi}{4}\right]$

This is the required solution.