#### Provide solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise  3.7 question 3 sub question (iv)

Answer:   $\frac{\pi }{4}$

Hint:  The range of principal value of $\tan ^{-1}$ is $\left [-\frac{\pi }{2} ,\frac{\pi }{2}\right ]$
Given: $\tan ^{-1}\left(\tan \frac{9 \pi}{4}\right)$

Explanation:

First we solve $\tan \frac{9 \pi}{4}$

$\tan \frac{9 \pi}{4}=\tan \left(2 \pi+\frac{\pi}{4}\right)$

$\therefore \tan (2 \pi+\theta)=\tan \theta$

\begin{aligned} &\tan \left(2 \pi+\frac{\pi}{4}\right)=\tan \frac{\pi}{4} \\ &\therefore \tan \frac{\pi}{4}=1 \\ &\tan \frac{9 \pi}{4}=1 \end{aligned}

By substituting this value in $\tan ^{-1}\left(\tan \frac{9 \pi}{4}\right)$, we get

$\tan ^{-1}(1)$

let,        $\tan ^{-1}(1)=y$

\begin{aligned} &\tan y=1 \\ &\tan \left(\frac{\pi}{4}\right)=1 \end{aligned}

The range of principal value of $\tan ^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text { and } \tan \left(\frac{\pi}{4}\right)=1$

$\tan ^{-1}\left(\tan \frac{9 \pi}{4}\right)=\frac{\pi}{4}\; \; \; \; \; \; \; \; \quad \tan ^{-1}(\tan x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$