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