Question

If ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\frac{1}{a}{\tan }^{-1}x$.
Find the value of a.

• A. $a=1/4$
• B. $a=1/2$
• C. $a=1$
• D. $a=2$

Answer 1

The correct option is D $a=2$

We have, ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\frac{1}{a}{\tan }^{-1}x$

Putting $x=\tan \theta$, we get
${\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)={\tan }^{-1}\left[\frac{\sqrt{1+{\tan }^2\theta }-1}{\tan \theta }\right]$
$={\tan }^{-1}\left[\frac{\sec \theta -1}{\tan \theta }\right]$
$={\tan }^{-1}\left[\frac{1-\cos \theta }{\sin \theta }\right]$

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