Question

Simplify : ${\tan }^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right]$

Answer 1

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

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

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

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

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

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

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

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

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

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

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