Question

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

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

Answer 1

The correct option is A $\frac{1-2x}{2\sqrt{(1-x^2)}}$

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

$=(\frac{\pi }{4}-\frac{\theta }{2})+\sin \left(2.\frac{\theta }{2}\right)$

$=\pi /4-\theta /2+\sqrt{1-{\cos }^2\theta }$

$=\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x+\sqrt{1-x^2}$

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