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  4. If operatornameTan-1((√1+x2-√1-x2/√1+x2+√1-x2))=α then x2=

Q. If $\operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right)=\alpha$ then $x^{2}=$

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Solution:

Put $x^{2}=\sin 2 \propto$
Then $\sqrt{1+x^{2}} =\sqrt{1+\sin 2 \alpha}$
$=\sqrt{\cos ^{2} \alpha+\sin ^{2} \alpha+2 \sin \alpha \cos \alpha}=\cos \alpha+\sin \alpha$
$\sqrt{1-x^{2}} =\cos \alpha-\sin \alpha$
$\therefore L H S=\tan ^{-1}\left[\frac{\cos \alpha+\sin \alpha-\cos \alpha+\sin \alpha}{\cos \alpha+\sin \alpha+\cos \alpha-\sin \alpha}\right]=\tan ^{-1}\left(\frac{2 \sin \alpha}{2 \cos \alpha}\right)$
$=\tan ^{-1}(\tan \alpha)=\alpha$

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