provide solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 7 sub question (x)

Answer:  $\sqrt{1-x^{2}}$

Hint:  The range of principal value of $\sin ^{-1}$ is $\left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]$

Given:  $\sin \left[2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right]$

Explanation:

First we solve  $2 \tan ^{-} \sqrt{\frac{1-x}{1+x}}$

Let            $x=\cos 2 A$

Therefore,

$2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}=2 \tan ^{-1} \sqrt{\frac{1-\cos 2 A}{1+\cos 2 A}}$

$\left[1-\cos 2 x=2 \sin ^{2} x \& 1+\cos 2 A=2 \cos ^{2} A\right]$

$=2 \tan ^{-1} \sqrt{\frac{2 \sin ^{2} A}{2 \cos ^{2} A}}$

\begin{aligned} &=2 \tan ^{-1} \sqrt{\tan ^{2} A} \\ &=2 \tan ^{-1}(\tan A) \end{aligned}

As we know,  $\tan ^{-1}(\tan x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

$=2 \tan ^{-1}(\tan A)=2 A$

Therefore,

$\sin \left[2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right]$

$=\sin 2 A \quad \therefore\left(1-\sin ^{2} \theta=\cos ^{2} \theta\right)$

$=\sqrt{1-\cos ^{2} 2 A}$

Now put value of $\cos 2 A=x$

$\sqrt{1-\cos ^{2} 2 A}=\sqrt{1-x^{2}}$