#### Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 7 sub question (ix) maths textbook solution

Answer:  $\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x$

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

Given:  $\sin ^{-1}\left[\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right], 0

Explanation:

Let         $x=\cos \theta$

Then

$\theta=\cos ^{-1} x$

Now,

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

$\therefore 1+\cos x=2 \cos ^{2}\left(\frac{x}{2}\right) \& 1-\cos x=2 \sin ^{2}\left(\frac{x}{2}\right)$

\begin{aligned} &\sin ^{-1}\left\{\frac{\sqrt{2 \cos ^{2} \frac{\theta}{2}}+\sqrt{2 \sin ^{2} \frac{\theta}{2}}}{2}\right\} \\ &\sin ^{-1}\left\{\frac{\sqrt{2} \cos \frac{\theta}{2}+\sqrt{2} \sin \frac{\theta}{2}}{2}\right\} \end{aligned}

\begin{aligned} &\sin ^{-1}\left\{\frac{\sqrt{2}\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)}{2}\right\} \\ &\therefore\{\sqrt{2} \times \sqrt{2}=2\} \end{aligned}

\begin{aligned} &\sin ^{-1}\left\{\frac{\sqrt{2}\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)}{\sqrt{2} \times \sqrt{2}}\right\} \\ &\sin ^{-1}\left\{\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\sqrt{2}}\right\} \\ &\sin ^{-1}\left\{\frac{1}{\sqrt{2}} \sin \frac{\theta}{2}+\frac{1}{\sqrt{2}} \cos \frac{\theta}{2}\right\} \end{aligned}

\begin{aligned} &\sin ^{-1}\left\{\sin \left(\frac{\theta}{2}+\frac{\pi}{4}\right)\right\} \\ &\sin ^{-1}(\sin \theta)=\theta, \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \end{aligned}

\begin{aligned} &=\frac{\theta}{2}+\frac{\pi}{2} \\ &=\frac{\cos ^{-1} x}{2}+\frac{\pi}{4} \\ &\sin ^{-1}\left\{\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right\}=\frac{\cos ^{-1} x}{2}+\frac{\pi}{4} \end{aligned}