The simplest form of $\tan ^{-1} \frac{\sqrt{1+x^2}-1}{x}, x \neq 0$ is:

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

$y=\tan ^{-1} \frac{\sqrt{1+x^2}-1}{x}$

let  $x = \tan \theta$       (substituting in y)   $\Rightarrow \quad \theta=\tan ^{-1}(x)$       .......(1)

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

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

since $1+\tan ^2 \theta=\sec ^2 \theta$

$y=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)$

as  $\sec \theta = \frac{1}{\cos \theta}$

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

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

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

multiplying and dividing by cos θ

$y=\tan ^{-1}\left(\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)$

as  $\cos \theta=1-2 \sin ^2 \frac{\theta}{2}$

$\Rightarrow 2 \sin ^2 \frac{\theta}{2}=1-\cos \theta$

$\sin \theta= 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$

$y=\tan ^{-1}\left(\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\right)$

$y=\tan ^{-1} \tan \frac{\theta}{2}=\frac{\theta}{2}$

from (1)

$y=\frac{1}{2} \tan ^{-1}(x)$