#### Please solve RD Sharma Class 12 Chapter Inverse Trigonometric Functions Exercise 3.3 Question 1 Subquestion (i) 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   $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=y$                                                                      .....(i)

$\tan y=\frac{1}{\sqrt{3}} \quad\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \left[\because \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]$

$\tan y=\tan \frac{\pi}{6}$

\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 $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

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

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