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

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

Hint: The range of principal value of  $\sec ^{-1}$ is    $\left [ 0,\pi \right ]-\left [ \frac{\pi }{2} \right ]$
Given:   $\sec ^{-1}\left(\sec \frac{25 \pi}{6}\right)$

Explanation:

First we solve $\sec \frac{25 \pi}{6}$

$\sec \left(\frac{25 \pi}{6}\right)=\sec \left(4 \pi+\frac{\pi}{6}\right)$

As we know   $\sec [2 \pi+\theta]=\sec \theta$

$\sec \left(4 \pi+\frac{\pi}{6}\right)=\sec \left(\frac{\pi}{6}\right)$

As we know  $\sec \left(\frac{\pi}{6}\right)=\frac{2 \sqrt{3}}{3}$

By substituting these value in $\sec ^{-1}\left(\sec \frac{25 \pi}{6}\right)$  we get,

$\sec ^{-1}\left(\frac{2 \sqrt{3}}{3}\right)$

Now,     $\text { let } y=\sec ^{-1}\left(\frac{2 \sqrt{3}}{3}\right)$

\begin{aligned} &\sec y=\frac{2 \sqrt{3}}{3} \\ &\sec \left(\frac{\pi}{6}\right)=\frac{2 \sqrt{3}}{3} \end{aligned}

The range of principal value of   $\sec ^{-1} \text { is }[0, \pi]-\left\{\frac{\pi}{2}\right\} \text { and } \sec \left(\frac{\pi}{6}\right)=\frac{2 \sqrt{3}}{3}$

\begin{aligned} &\sec ^{-1}\left(\sec \frac{25 \pi}{6}\right)=\frac{\pi}{6} \\ &\sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\} \end{aligned}