Question

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

Answer 1

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

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

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

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

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

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

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