Question

Simplify : ${\cot }^{-1}[\sqrt{1+x^2}-x]$

Answer 1

${\cot }^{-1}[\sqrt{1+x^2}-x]$

Let $x=\cot \theta$

${\cot }^{-1}[\sqrt{1+{\cot }^2\theta }-\cot \theta ]$

$={\cot }^{-1}[\text{cosec}\theta -\cot \theta ]$

$={\cot }^{-1}[\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta }]$

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

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

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

$={\cot }^{-1}\left(\cot (\frac{\pi }{2}-\frac{\theta }{2})\right)$

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

$=\frac{\pi }{2}-\frac{{\cot }^{-1}x}{2}$