If
Question:

If $\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right)=α, then $x^{2}=$

(a) $\sin 2 \alpha$

(b) $\sin a$

(c) $\cos 2 a$

(d) $\cos \alpha$

Solution:

(a) $\sin 2 \alpha$

$\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right)=\alpha$

$\Rightarrow \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}=\tan \alpha$

$\Rightarrow \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}} \times \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}=\tan \alpha$

$\Rightarrow \frac{\left(\sqrt{1+x^{2}}\right)^{2}+\left(\sqrt{1-x^{2}}\right)^{2}-2 \sqrt{1+x^{2}} \sqrt{1-x^{2}}}{\left(\sqrt{1+x^{2}}\right)^{2}-\left(\sqrt{1-x^{2}}\right)^{2}}=\tan \alpha$

$\Rightarrow \frac{1-\sqrt{1-x^{4}}}{x^{2}}=\tan \alpha$

$\Rightarrow x^{2} \tan \alpha=1-\sqrt{1-x^{4}}$

$\Rightarrow \sqrt{1-x^{4}}=1-x^{2} \tan \alpha$

$\Rightarrow 1-x^{4}=1+x^{4} \tan ^{2} \alpha-2 x^{2} \tan \alpha$

$\Rightarrow x^{4}+x^{4} \tan ^{2} \alpha-2 x^{2} \tan \alpha=0$

$\Rightarrow x^{4} \sec ^{2} \alpha-2 x^{2} \tan \alpha=0$

$\Rightarrow x^{2}\left(x^{2} \sec ^{2} \alpha-2 \tan \alpha\right)=0$

$\Rightarrow x^{2} \sec ^{2} \alpha-2 \tan \alpha=0 \quad\left[\because x^{2} \neq 0\right]$

$\Rightarrow x^{2} \sec ^{2} \alpha=2 \tan \alpha$

$\Rightarrow x^{2}=\frac{2 \tan \alpha}{\sec ^{2} \alpha}=2 \sin \alpha \cos \alpha=\sin 2 \alpha$

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