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 $\frac{d y}{d x}=$

• A. $\frac{1-2 x}{2 \sqrt{1-x^{2}}}$
• B. $\frac{1-2 x}{x \sqrt{1-x^{2}}}$
• C. $\frac{2 x+1}{x \sqrt{1-x}}$
• 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
$y=\tan ^{-1}\left\{\frac{\cos 2 \theta}{1+\sqrt{\sin ^{2} 2 \theta}}\right\}$
$+\sin \left\{2 \tan ^{-1} \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}\right\}$
$\Rightarrow y=\tan ^{-1}\left\{\frac{\cos 2 \theta}{1+\sin 2 \theta}\right\}$
$+\sin \left\{2 \tan ^{-1} \sqrt{\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}}\right\}$
$\Rightarrow y=\tan ^{-1}\left\{\frac{\cos ^{2} \theta-\sin ^{2} \theta}{(\cos \theta+\sin \theta)^{2}}\right\}$
$+\sin \left\{2 \tan ^{-1}(\tan \theta)\right\}$
$\Rightarrow y=\tan ^{-1}\left\{\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\right\}+\sin 2 \theta$
$\Rightarrow y=\tan ^{-1}\left\{\frac{1-\tan \theta}{1+\tan \theta}\right\}+\sin 2 \theta$
$\left.\Rightarrow \tan \left(\frac{\pi}{4}-\theta\right)\right)+\sin 2 \theta$
$\Rightarrow y=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\theta\right)\right)+\sin 2 \theta$
$\Rightarrow y=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x+\sqrt{1-x^{2}}$
$\left[\because \sin 2 \theta=\sqrt{\left.1-\cos ^{2} 2 \theta\right]}\right.$
Now, on differentiating both sides w.r.t. $x$, we get
$\frac{d y}{d x} =\frac{1}{2} \cdot \frac{1}{\sqrt{1-x^{2}}}+\frac{1}{2 \sqrt{1-x^{2}}}(-2 x)$
$=\frac{1-2 x}{2 \sqrt{1-x^{2}}}$