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

Answer : $\frac{-\pi }{6}$

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

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

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

Explanation :

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

$\tan y=\frac{-1}{\sqrt{3}}$

$\tan y=-\tan \left(\frac{\pi}{6}\right)$

$\tan y=\tan \left(\frac{-\pi}{6}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \left [ \because \tan (-\theta)=-\tan \theta \right ]$

\begin{aligned} &y=\frac{-\pi}{6} \\ &\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-\pi}{6} \end{aligned}                                                                   [From equation (i)]

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

$\because \quad \tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

Hence, principal value of  $\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-\pi}{6}$.