If $ an ^{-1}\left[\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}} ight]=\alpha$, then the value of $\sin 2 \alpha$ is

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If $	an ^{-1}\left[\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}ight]=\alpha$, then the value of $\sin 2 \alpha$ is

Aicharya · Accepted Answer

We have $	an \alpha=\left[\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}ight]$ Put $x^2=\cos 	heta$ $\begin{aligned} & 	herefore 	an \alpha=\frac{\sqrt{2 \cos ^2 \frac{	heta}{2}-\sqrt{2 \sin ^2 \frac{	heta}{2}}}}{\sqrt{2 \cos ^2 \frac{	heta}{2}+\sqrt{2 \sin ^2 \frac{	heta}{2}}}} \ & 	herefore 	an \alpha=\frac{\cos \frac{	heta}{2}-\sin \frac{	heta}{2}}{\cos \frac{	heta}{2}-\sin \frac{	heta}{2}}=\frac{1-	an \frac{	heta}{2}}{1+	an \frac{	heta}{2}}=	an \left(\frac{\pi}{4}-\frac{	heta}{2}ight) \ & 	herefore \alpha=\frac{\pi}{4}-\frac{	heta}{2} \Rightarrow 2 \alpha=\frac{\pi}{2}-	heta \ & 	herefore \sin 2 \alpha=\sin \left(\frac{\pi}{2}-	hetaight)=\cos 	heta=x^2 \end{aligned}$

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…

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