Question

If $y={\tan }^{-1}\frac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}$, then $\frac{dy}{dx}$ is equal to.

• A. $\frac{-1}{\sqrt{1-x^2}}$
• B. $\frac{1}{\sqrt{1-x^2}}$
• C. $\frac{-x}{\sqrt{1-x^2}}$
• D. $Noneofthese$

Answer 1

The correct option is B $\frac{1}{\sqrt{1-x^2}}$

Formula:

${\tan }^{-1}\left(\tan x\right)=x$

Let, $x=\cos \theta ⟹\theta ={\cos }^{-1}x$

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

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

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

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

$y={\tan }^{-1}\left[\tan \right(\frac{\pi }{4}-\theta \left)\right]$

$y=\frac{\pi }{4}-\theta$

$y=\frac{\pi }{4}-{\cos }^{-1}x$

$\frac{dy}{dx}=0-\frac{-1}{\sqrt{1-x^2}}$

$\therefore \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$