#### explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 7 sub question (viii) maths

Answer:  $\frac{\pi}{4}+\sin ^{-1} x$

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

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

Explanation:

Let         $x=\sin \alpha$

Then      $\alpha=\sin ^{-1} x$

\begin{aligned} &\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}=\frac{\sin \alpha+\sqrt{1-\sin ^{2} \alpha}}{\sqrt{2}} \\ &\therefore \sin ^{2} \alpha+\cos ^{2} \alpha=1 \end{aligned}

Then,     $1-\sin ^{2} \alpha=\cos ^{2} \alpha$

\begin{aligned} &=\frac{\sin \alpha+\sqrt{\cos ^{2} \alpha}}{\sqrt{2}} \\ &=\frac{\sin \alpha+\sqrt{\cos ^{2} \alpha}}{\sqrt{2}} \end{aligned}

\begin{aligned} &=\left(\sin \alpha \cdot \frac{1}{\sqrt{2}}+\cos \alpha \cdot \frac{1}{\sqrt{2}}\right) \\ &\therefore \cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}, \sin \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \end{aligned}

\begin{aligned} &=\sin \alpha \cdot \cos \left(\frac{\pi}{4}\right)+\cos \alpha \cdot \sin \left(\frac{\pi}{4}\right) \\ &\therefore \sin A \cos B+\cos A \sin B=\sin (A+B) \end{aligned}

$\sin \alpha \cos \left(\frac{\pi}{4}\right)+\cos \alpha \sin \left(\frac{\pi}{4}\right)=\sin \left(\alpha+\frac{\pi}{4}\right)$

Then put value of    $\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}=\sin \left(\alpha+\frac{\pi}{4}\right)$

$\sin ^{-1}\left(\sin \left(\alpha+\frac{\pi}{4}\right)\right)$

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

\begin{aligned} &=\alpha+\frac{\pi}{4} \\ &=\sin ^{-1} x+\frac{\pi}{4} \end{aligned}