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

Answer:   $\frac{\pi}{2}-\frac{1}{2} \cot ^{-1} x$

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

Given:  $\tan ^{-1}\left\{x+\sqrt{1+x^{2}}\right\}, x \in R$

Explanation:

Let         $x=\cot \theta ; \theta=\cot ^{-1} x$

Now ,

$\tan ^{-1}\left\{x+\sqrt{1+x^{2}}\right\}=\tan ^{-1}\left\{\cot \theta+\sqrt{1+\cot ^{2} \theta}\right\}$

As we know,  $1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta$

\begin{aligned} &\tan ^{-1}\left\{\cot \theta+\sqrt{\operatorname{cosec}^{2} \theta}\right\} \\ &\tan ^{-1}\{\cot \theta+\operatorname{cosec} \theta\} \end{aligned}

As we know,  $\cot \theta=\frac{\cos \theta}{\sin \theta^{\prime}} \sin \theta=\frac{1}{\operatorname{cosec} \theta}$

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

As we know, $\sin 2 \theta=2 \sin \theta \cos \theta$

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

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

As we know $\tan \theta=\cot \left(\frac{\pi}{2}-\theta\right)$

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

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

$\Rightarrow \quad\left(\frac{\pi}{2}-\frac{\theta}{2}\right)$

Now put value of θ

$\Rightarrow \quad \frac{\pi}{2}-\frac{\cot ^{-1} x}{2}$