#### Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 3 sub question (iii) maths textbook solution

Answer:   $\frac{\pi }{6}$

Hint:  The range of principal value of $\tan ^{-1}$ is $\left [-\frac{\pi }{2} ,\frac{\pi }{2}\right ]$
Given: $\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)$

Explanation:

First we solve $\tan \frac{7 \pi}{6}$

$\tan \frac{7 \pi}{6}=\tan \left(\pi+\frac{\pi}{6}\right)$

As we know, $\tan (\pi+\theta)=\tan \theta$

\begin{aligned} &\tan \left(\pi+\frac{\pi}{6}\right)=\tan \frac{\pi}{6} \\ &\therefore \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \end{aligned}

By substituting this value in $\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)$ we get

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

Now,     $\text { let } \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=y$

\begin{aligned} &\tan y=\frac{1}{\sqrt{3}} \\ &\tan \left(\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}} \end{aligned}

The range of principal value of  $\tan ^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text { and } \tan \left(\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}}$

$\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=\frac{\pi}{6} \quad \text { As } \quad \tan ^{-1}(\tan x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$