Question

Writhe the following functions in the simplest form -
${\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}$,$x\ne 0$

Answer 1

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

Let $x=\tan \theta$

$\Rightarrow {\tan }^{-1}(\frac{\sqrt{1+{\tan }^2\theta }-1}{\tan \theta }$

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

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

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

$\Rightarrow {\tan }^{-1}\left(\tan \frac{\theta }{2}\right)$

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