Question

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

Answer 1

let

$x=\tan a\Rightarrow a={\tan }^{-1}x$

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

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

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

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

$={\tan }^{-1}\left(\frac{1-cosa}{-sina}\right)$

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

$={\tan }^{-1}(-\tan \frac{a}{2})=\frac{a}{2}$

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