#### Please solve RD Sharma Class 12 Chapter Inverse Trigonometric Functions Exercise 3.3 Question 1 Subquestion (iv) maths textbook solution.

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

Hint :  The branch with range $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is called the principal value branch of function $\tan^{-1}$

Thus, $\tan ^{-1}: R \rightarrow\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

Given : $\tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)$

Explanation :

Let $y=\tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)$                                                                                                  ....(i)

$y=\tan ^{-1}\left[2\left(\frac{-1}{2}\right)\right]$                                                                                                  $\left [ \because \cos\frac{2\pi }{3}=\frac{-1}{2} \right ]$

\begin{aligned} &y=\tan ^{-1}(-1) \\ &\tan y=-1 \end{aligned}

$\tan y=-\tan \left(\frac{\pi}{4}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \left [ \because \tan (-\theta)=-\tan \theta \right ]$

$y=\frac{-\pi}{4}$

$\tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)=\frac{-\pi}{4}$

The range of principal value branch of $\tan^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.

$\because \quad \tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)=\frac{-\pi}{4} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

Hence, the principal value of $\tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)$ is $\frac{-\pi}{4}$.