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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.

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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.

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