#### Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 4 sub question (vii) maths textbook solution

Answer: $\frac{3\pi }{4}$

Hint: The range of principal value of  $\sec ^{-1}$ is $\left [ 0,\pi \right ]-\left [ \frac{\pi }{2} \right ]$
Given:  $\sec ^{-1}\left(\sec \frac{13 \pi}{4}\right)$

Explanation:

First we solve  $\sec \left(\frac{13 \pi}{4}\right)$

$\sec \left(\frac{13 \pi}{4}\right)=\sec \left(4 \pi-\frac{3 \pi}{4}\right)$

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

But here after subtract $\frac{3\pi }{4}$   it comes in III quadrant where sec is negative

\begin{aligned} &\sec \left(4 \pi-\frac{3 \pi}{4}\right)=-\sec \left(\frac{3 \pi}{4}\right) \\ &\therefore \sec \frac{3 \pi}{4}=\sqrt{2} \\ &-\sec \left(\frac{3 \pi}{4}\right)=-\sqrt{2} \end{aligned}

By substituting these values in $\sec ^{-1}\left(\sec \frac{13 \pi}{4}\right)$ we get,

$\sec ^{-1}(-\sqrt{2})$

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

\begin{aligned} &\sec y=-\sqrt{2} \\ &-\sec y=\sqrt{2} \\ &-\sec \left(\frac{\pi}{4}\right)=\sqrt{2} \\ &\sec \left(\pi-\frac{\pi}{4}\right)=\sqrt{2} \\ &\sec \left(\frac{3 \pi}{4}\right)=-\sqrt{2} \end{aligned}

The range of the principle value of   $\sec ^{-1} \text { is }[0, \pi]-\left\{\frac{\pi}{2}\right\} \text { and } \sec \left(\frac{3 \pi}{4}\right)=-\sqrt{2}$

$\sec ^{-1}\left(\sec \frac{3 \pi}{4}\right)=\frac{3 \pi}{4}$

$\therefore \sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\}$