#### need solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 4 sub question (v)

Answer:  $\frac{\pi }{5}$

Hint: The range of principal value of  $\sec ^{-1}$ is $\left [ 0,\pi \right ]-\left [ \frac{\pi }{2} \right ]$
Given:  $\sec ^{-1}\left(\sec \frac{9 \pi}{5}\right)$

Explanation:

First we solve $\sec \left(\frac{9 \pi}{5}\right)$

\begin{aligned} &\sec \left(\frac{9 \pi}{5}\right)=\sec \left(2 \pi-\frac{\pi}{5}\right) \\ &\therefore[\sec (2 \pi-\theta)]=\sec \theta \end{aligned}

\begin{aligned} &\sec \left(2 \pi-\frac{\pi}{5}\right)=\sec \left(\frac{\pi}{5}\right) \\ &\sec \left(\frac{9 \pi}{5}\right)=\sec \left(\frac{\pi}{5}\right) \end{aligned}

By substituting these value in $\sec ^{-1}\left(\sec \frac{9 \pi}{5}\right)$ we get,

\begin{aligned} &\sec ^{-1}\left(\sec \frac{\pi}{5}\right) \\ &\therefore \sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\} \\ &\sec ^{-1}\left(\sec \frac{\pi}{5}\right)=\frac{\pi}{5} \end{aligned}

Hence, $\sec ^{-1}\left(\sec \frac{9 \pi}{5}\right)=\frac{\pi}{5}$