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

Answer:  $\frac{1}{2} \cot ^{-1} x$

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

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

Explanation:

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

Now,

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

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

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

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

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

$\\ \\ \hspace{0.5cm}\because 1-\cos x=2\sin^2\frac{x}{2}\\ \\ \hspace{0.5cm}\because \sin x=2sin \frac{x}{2} \cdot \cos \frac{x}{2}$

\begin{aligned} &\tan ^{-1}\left\{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta }{2}}\right\} \\ &\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, the principal range of $\tan ^{-1}$ is $\left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]$

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

$\begin{array}{ll} \Rightarrow & \frac{\theta}{2} \end{array}$

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