Question

Write ${\tan }^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right],x\ne 0$ in the simplest form.

Answer 1

Let $x=\tan \theta$
${\tan }^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right]={\tan }^{-1}\left[\frac{\sec \theta -1}{\frac{\sin \theta }{\cos \theta }}\right]$
$={\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)$

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

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

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