$\begin{aligned} & \text { If } y=\tan ^{-1}\left\{\frac{x}{1+\sqrt{1-x^2}}\right\} \\ & +\sin \left\{2 \tan…

$\begin{aligned} & \text { If } y=\tan ^{-1}\left\{\frac{x}{1+\sqrt{1-x^2}}\right\} \\ & +\sin \left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\} \text {, then } \frac{d y}{d x}=\end{aligned}$
  1. $\frac{1-2 x}{2 \sqrt{1-x^2}}$
  2. $\frac{1-2 x}{x \sqrt{1-x^2}}$
  3. $\frac{2 x+1}{x \sqrt{1-x}}$
  4. $\frac{2-x}{2 \sqrt{1-x^2}}$

Solution

Given that, $ y=\tan ^{-1}\left\{\frac{x}{1+\sqrt{1-x^2}}\right\}+\sin \left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\} $ Let $x=\cos 2 \theta$ $ \begin{aligned} & 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 \\ & y=\tan ^{-1}\left\{\frac{\cos \theta\left(1-\frac{\sin \theta}{\cos \theta}\right)}{\cos \theta\left(1+\frac{\sin \theta}{\cos \theta}\right)}\right\}+\sin 2 \theta \\ & \end{aligned} $ $ \begin{aligned} & \Rightarrow \quad y=\tan ^{-1}\left\{\frac{1-\tan \theta}{1+\tan \theta}\right\}+\sin 2 \theta \\ & \Rightarrow \quad y=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\theta\right)\right)+\sin 2 \theta \\ & \Rightarrow \quad y=\frac{\pi}{4}-\theta+\sqrt{1-\cos ^2} 2 \theta \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{1}{2} \cdot \frac{1}{\sqrt{1-r^2}}+\frac{1}{2 \sqrt{1-r^2}}(2 x) \end{aligned} $ Now, on differentiating w.r.t. $x$, we get $ \begin{aligned} & \frac{d y}{d x}=\frac{1}{2} \cdot \frac{1}{\sqrt{1-x^2}}+\frac{1}{2 \sqrt{1-x^2}}(-2 \mathrm{x}) \\ & \therefore \quad \frac{d y}{d x}=\frac{1-2 x}{2 \sqrt{1-x^2}} \end{aligned} $

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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