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

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

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

Given:  $\tan ^{-1} \sqrt{\frac{a-x}{a+x}},-a

Explanation:

Let         $x=a \cos \theta$

then        $\theta=\cos ^{-1} \frac{x}{a}$

Now,

$\tan ^{-1} \sqrt{\frac{a-x}{a+x}}=\tan ^{-1} \sqrt{\frac{a-a \cos \theta}{a+a \cos \theta}}$

\begin{aligned} &\tan ^{-1} \sqrt{\frac{a(1-\cos \theta)}{a(1+\cos \theta)}} \\ &\tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \end{aligned}

$\tan ^{-1} \sqrt{\frac{2 \sin \frac{2}{2}}{2 \cos ^{2} \frac{\theta}{2}}} \quad\left[\begin{array}{l} \left.\because \begin{array}{l} 1+\cos x=2 \cos ^{2}\left(\frac{x}{2}\right) \\ 1-\cos x=2 \sin ^{2}\left(\frac{x}{2}\right) \end{array}\right] \end{array}\right.$

$\tan ^{-1} \sqrt{\tan ^{2} \frac{\theta}{2}} \quad \because \frac{\sin \theta}{\cos \theta}=\tan \theta$

$\tan ^{-1}\left(\tan \frac{\theta}{2}\right)$

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} \cos ^{-1} \frac{x}{a} \end{aligned}