Question

If $y={\tan }^{-1}\left\{\frac{x}{1+\sqrt{1-x^2}}\right\}$
$+\sin \left\{2{\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right\}$, then $\frac{dy}{dx}=$

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

Answer 1

Answer: (A)

We have,
$y={\tan }^{-1}\left\{\frac{x}{1+\sqrt{1-x^2}}\right\}+\sin \left\{2{\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right\}$
On substituting $x=\cos 2\theta$, we get
$y={\tan }^{-1}\left\{\frac{\cos 2\theta }{1+\sqrt{{\sin }^{2}2\theta }}\right\}$
$+\sin \left\{2{\tan }^{-1}\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}\right\}$
$\Rightarrow y={\tan }^{-1}\left\{\frac{\cos 2\theta }{1+\sin 2\theta }\right\}$
$+\sin \left\{2{\tan }^{-1}\sqrt{\frac{2{\sin }^2\theta }{2{\cos }^2\theta }}\right\}$
$\Rightarrow y={\tan }^{-1}\left\{\frac{{\cos }^2\theta -{\sin }^2\theta }{(\cos \theta +\sin \theta {)}^2}\right\}$
$+\sin \left\{2{\tan }^{-1}(\tan \theta )\right\}$
$\Rightarrow y={\tan }^{-1}\left\{\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }\right\}+\sin 2\theta$
$\Rightarrow y={\tan }^{-1}\left\{\frac{1-\tan \theta }{1+\tan \theta }\right\}+\sin 2\theta$
$\Rightarrow \tan \left(\frac{\pi }{4}-\theta \right))+\sin 2\theta$
$\Rightarrow y={\tan }^{-1}\left(\tan \left(\frac{\pi }{4}-\theta \right)\right)+\sin 2\theta$
$\Rightarrow y=\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x+\sqrt{1-x^2}$
$[\because \sin 2\theta =\sqrt{1-{\cos }^{2}2\theta ]}$
Now, on differentiating both sides w.r.t. $x$, we get
$\frac{dy}{dx}=\frac{1}{2}\cdot \frac{1}{\sqrt{1-x^2}}+\frac{1}{2\sqrt{1-x^2}}(-2x)$
$=\frac{1-2x}{2\sqrt{1-x^2}}$