#### Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 4 sub question (iii) 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:  $\sec ^{-1}\left(\sec \frac{5 \pi}{4}\right)$

Explanation:

First we solve  $\sec \frac{5 \pi}{4}$

But $\left(2 \pi-\frac{3 \pi}{4}\right)$  belongs to 3 quadrant

So, $\sec \left(2 \pi-\frac{3 \pi}{4}\right)=\sec \frac{3 \pi}{4}$

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

By substituting these value in  $\sec ^{-1}\left(\sec \frac{5 \pi}{4}\right)$  we get,

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

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

$\Rightarrow \quad \sec y=-\sqrt{2}$

$\Rightarrow \quad-\sec \left(\frac{\pi}{4}\right)=\sqrt{2}$

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

The range of principal value of  $\sec ^{-1} \text { is }[0, \pi]-\left\{\frac{\pi}{2}\right\}$

$\sec ^{-1}\left(\sec \frac{3 \pi}{4}\right)=\frac{3 \pi}{4} \; \; \; \; \; \; \; \; \; \; \quad \therefore \sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\}$