Question

If $y={\tan }^{-1}\frac{x}{1+\sqrt{1-x^2}}+\sin \left(2{\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right)$, then find $\frac{dy}{dx}$ for $x\in (-1,1)$.

Answer 1

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

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

Put $x=\cos 2\theta$

$\theta =\frac{1}{2}{\cos }^{-1}x$ ......(1)

The RHS becomes,

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

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

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

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

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

$={\tan }^{-1}\frac{(1-\tan \theta )}{(1+\tan \theta )}+\sin 2\theta$

$={\tan }^{-1}\frac{(\tan \frac{\pi }{4}-\tan \theta )}{(1+\tan \frac{\pi }{4}\tan \theta )}+\sin 2\theta$

$={\tan }^{-1}\tan (\frac{\pi }{4}-\theta )+\sin 2\theta$

$=\frac{\pi }{4}-\theta +\sin 2\theta$

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

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

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