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If $\mathrm{y}=\tan ^{-1}\left[\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}}\right]$, then $\left(\frac{d y}{d x}\right)=$
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The correct answer is:
$\frac{1}{\sqrt{1-x^{2}}}$
Given $y=\tan ^{-1}\left[\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}}\right]$
Put $x=\cos \theta \Rightarrow \theta=\cos ^{-1} x$
$y=\tan ^{-1}\left[\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\right]=\tan ^{-1}\left[\frac{1-\tan \theta}{1+\tan \theta}\right]$
$\quad=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\theta\right)\right]=\frac{\pi}{4}-\theta$
$\therefore \frac{d y}{d x}=\frac{\pi}{4}-\cos ^{-1} x$
$$
\begin{aligned} y &=\frac{1}{\sqrt{1-x^{2}}} \\ y &=\frac{1}{\sqrt{1-x^{2}}} \end{aligned}
$$
Put $x=\cos \theta \Rightarrow \theta=\cos ^{-1} x$
$y=\tan ^{-1}\left[\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\right]=\tan ^{-1}\left[\frac{1-\tan \theta}{1+\tan \theta}\right]$
$\quad=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\theta\right)\right]=\frac{\pi}{4}-\theta$
$\therefore \frac{d y}{d x}=\frac{\pi}{4}-\cos ^{-1} x$
$$
\begin{aligned} y &=\frac{1}{\sqrt{1-x^{2}}} \\ y &=\frac{1}{\sqrt{1-x^{2}}} \end{aligned}
$$
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