Question

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

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

Answer 1

Answer: (B)

Put $x=\cos \theta$
$y={\tan }^{-1}\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }={\tan }^{-1}\frac{1-\tan \theta }{1+\tan \theta }$
$={\tan }^{-1}\tan \left(\frac{\pi }{4}-\theta \right)$
$=\frac{\pi }{4}-\theta =\frac{\pi }{4}-{\cos }^{-1}x$
$\therefore \frac{dy}{dx}=0-\left(\frac{1}{\sqrt{1-x^2}}\right)=\frac{1}{\sqrt{1-x^2}}$