Question

Write ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)x\ne 0$ in the simplest form

Answer 1

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

Putting $x=\tan \theta$ ………. (1)

We get,

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

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

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

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

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

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

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

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

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

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

By equation (1) and we get,

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

Then,

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

Hence, this is the answer