Question

Prove that ${\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}=\frac{1}{2}{\tan }^{-1}x,x\ne 0$

Answer 1

To find the value of ${\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}$ let $x=\tan \theta$.....(1)

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

$={\tan }^{-1}\frac{\sec \theta -1}{\tan \theta }$

$={\tan }^{-1}\frac{1-\cos \theta }{\sin \theta }$

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

$=\frac{1}{2}\theta$

$=\frac{1}{2}{\tan }^{-1}x$ [Using (1)]