#### need solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 7 sub question (vii)

Answer:   $\frac{1}{2} \sin ^{-1} \frac{x}{a}$

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

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

Explanation:

Put        $x=a \sin \theta, \text { then } \theta=\sin ^{-1} \frac{x}{a}$

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

$\tan ^{-1}\left[\frac{a \sin \theta}{a+\sqrt{a^{2}\left(1-\sin ^{2} \theta\right)}}\right] \quad \because\left[\sin ^{2} \theta+\cos ^{2} \theta=1\right]$

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

\begin{aligned} &\tan ^{-1}\left[\frac{a \sin \theta}{a(1+\cos \theta)}\right] \\ &\tan ^{-1}\left[\frac{\sin \theta}{1+\cos \theta}\right] \end{aligned}

$\because \left[\begin{array}{l} \sin 2 \theta=2 \sin \theta \cos \theta \\ \cos 2 x=2 \cos ^{2} x-1 \end{array}\right]$

$\tan ^{-1}\left\{\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{1+2 \cos ^{2} \frac{\theta}{2}-1}\right\}$

\begin{aligned} &\tan ^{-1}\left[\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\right] \\ &\tan ^{-1}\left[\tan \frac{\theta}{2}\right] \end{aligned}

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

\begin{aligned} &\Rightarrow \quad \frac{\theta}{2} \\ &\Rightarrow \quad \frac{1}{2} \sin ^{-1} \frac{x}{a} \end{aligned}