Question

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

Answer 1

Given :
${\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}$
Substitute $x=\tan \theta$ in given equation

$\Rightarrow {\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}={\tan }^{-1}\left(\frac{\sqrt{1+{\tan }^2\theta }-1}{\tan \theta }\right)$
$\Rightarrow {\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}={\tan }^{-1}\left(\frac{\sqrt{{\sec }^2\theta }-1}{\tan \theta }\right)$
$\Rightarrow {\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)$
$\Rightarrow {\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}={\tan }^{-1}\left(\frac{\frac{1}{\cos \theta }-1}{\frac{\sin \theta }{\cos \theta }}\right)$
$\Rightarrow {\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}={\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)$
Using half angle formula
$\Rightarrow {\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}={\tan }^{-1}\left(\frac{2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\right)$
$\Rightarrow {\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}={\tan }^{-1}\left(\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}\right)$
$\Rightarrow {\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{\theta }{2}$
$\because x=\tan \theta$
$\Rightarrow \theta ={\tan }^{-1}x$
So,
$\frac{\theta }{2}=\frac{1}{2}{\tan }^{-1}x$