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If $y=\sin ^{2}\left(\tan ^{-1} \sqrt{\frac{1-x^{2}}{1+x^{2}}}\right)$, then $\frac{d y}{d x}=$
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The correct answer is:
$-x$
We have, $y=\sin ^{2}\left[\tan ^{-1} \sqrt{\frac{1-x^{2}}{1+x^{2}}}\right]$
$$
\begin{aligned}
\text{Put} \quad x^{2} &=\cos \theta \\
\text{So,} \quad y &=\sin ^{2}\left[\tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}\right] \\
&=\sin ^{2}\left[\tan ^{-1}\left(\tan \frac{\theta}{2}\right)\right] \\
&=\sin ^{2} \frac{\theta}{2}=\frac{1-\cos \theta}{2}=\frac{1-x^{2}}{2} \\
\frac{d y}{d x} &=\frac{-2 x}{2}=-x
\end{aligned}
$$
$$
\begin{aligned}
\text{Put} \quad x^{2} &=\cos \theta \\
\text{So,} \quad y &=\sin ^{2}\left[\tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}\right] \\
&=\sin ^{2}\left[\tan ^{-1}\left(\tan \frac{\theta}{2}\right)\right] \\
&=\sin ^{2} \frac{\theta}{2}=\frac{1-\cos \theta}{2}=\frac{1-x^{2}}{2} \\
\frac{d y}{d x} &=\frac{-2 x}{2}=-x
\end{aligned}
$$
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