If tan ^{-1}left[frac{sqrt{1+x^2}-sqrt{1-x^2} | vedclass

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(D) Given $	an \alpha = \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}$.Let $x^2 = \cos 	heta$. Then $	heta = \cos^{-1}(x^2)$.Substitu

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$x^2$

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(D) Given $	an \alpha = \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}$.Let $x^2 = \cos 	heta$. Then $	heta = \cos^{-1}(x^2)$.Substituting $x^2 = \cos 	heta$ into the expression:$	an \alpha = \frac{\sqrt{1+\cos 	heta}-\sqrt{1-\cos 	heta}}{\sqrt{1+\cos 	heta}+\sqrt{1-\cos 	heta}}$Using half-angle identities $1+\cos 	heta = 2\cos^2(	heta/2)$ and $1-\cos 	heta = 2\sin^2(	heta/2)$:$	an \alpha = \frac{\sqrt{2}\cos(	heta/2) - \sqrt{2}\sin(	heta/2)}{\sqrt{2}\cos(	heta/2) + \sqrt{2}\sin(	heta/2)}$Dividing numerator and denominator by $\sqrt{2}\cos(	heta/2)$:$	an \alpha = \frac{1 - 	an(	heta/2)}{1 + 	an(	heta/2)} = 	an(\frac{\pi}{4} - \frac{	heta}{2})$Thus,$\alpha = \frac{\pi}{4} - \frac{	heta}{2}$,which implies $2\alpha = \frac{\pi}{2} - 	heta$.Therefore,$\sin 2\alpha = \sin(\frac{\pi}{2} - 	heta) = \cos 	heta$.Since $x^2 = \cos 	heta$,we have $\sin 2\alpha = x^2$.

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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