Question

Simplify: ${\cot }^{-1}(\sqrt{1+x^2}+x).=\frac{{\csc }^{-1}x}{m}$.Find $m$

Answer 1

${\cot }^{-1}(\sqrt{1+x^2}+x)$

Let $x=\csc \theta$

$\Rightarrow \cot -1(\sqrt{1+{\csc }^2\theta }+\csc \theta )$

$\Rightarrow {\cot }^{-1}(\sqrt{{\cot }^2\theta }+\csc \theta )$

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

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

$\Rightarrow {\cot }^{-1}\left[\cot \frac{\theta }{2}\right]=\frac{\theta }{2}=\frac{1}{2}{\csc }^{-1}x$

comparing with $\frac{{\csc }^{-1}x}{m}$

$\therefore m=2$