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

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

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

Given: $\tan ^{-1}\left\{\frac{\sqrt{1+x^{2}}-1}{x}\right\}, x \neq 0$

Explanation:

Let       $x=\tan \theta, \text { then } \theta=\tan ^{-1} x$

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

As we know, $1+\tan ^{2} \theta=\sec ^{2} \theta$

\begin{aligned} &\tan ^{-1}\left(\frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\right) \\ &\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) \end{aligned}

$\tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right) \quad \because\left[\sec \theta=\frac{1}{\cos \theta}, \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$

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

$\tan ^{-1}\left(\frac{1-\cos \theta}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}\right)$

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

$\tan ^{-1}\left[\frac{2 \sin^{2} \frac{ \theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right] \quad \because\left[\begin{array}{c} \sin 2 \theta=2 \sin \theta \cos \theta \\ 1-\cos \theta=2 \sin ^{2} \frac{\theta}{2} \end{array}\right]$

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

$\tan ^{-1}\left[\tan \frac{\theta}{2}\right]=\frac{\theta}{2}$

Now putting the value of θ

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