need solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 2 sub question (i)

Answer:   $\frac{\pi }{4}$

Hint: The principal value branch of function  $\cos ^{-1}$  is  $\left [ 0,\pi \right ]$

Given:   $\cos ^{-1}\left\{\cos \left(-\frac{\pi}{4}\right)\right\}$

Explanation:

We know that $\cos \left(-\frac{\pi}{4}\right)=\cos \left(\frac{\pi}{4}\right) \quad \because[\cos (-\theta)=\cos \theta]$

Also know that, $\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$

By substituting these values in $\cos ^{-1}\left\{\cos \left(-\frac{\pi}{4}\right)\right\}$

we get,

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

Let        $y=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)$

$\cos y=\frac{1}{\sqrt{2}}$

Hence, range of principal value of $\cos ^{-1} \text { is }[0, \pi] \text { and } \cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$

\begin{aligned} &\cos ^{-1}\left\{\cos \left(-\frac{\pi}{4}\right)\right\}=\frac{\pi}{4} \\ &\therefore \cos ^{-1}(\cos x)=x, x \in[0, \pi] \end{aligned}