Question

If ${\tan }^{-1}\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}=\alpha$, then $x^2=$

• A. $\cos 2\alpha$
• B. $\sin 2\alpha$
• C. $\tan 2\alpha$
• D. $\cot 2\alpha$

Answer 1

Answer: (B)

$\sin 2\alpha$

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

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

Using componendo and dividendo, we get

$\frac{2\sqrt{1+x^2}}{2\sqrt{1-x^2}}=\frac{1+\tan \alpha }{1-\tan \alpha }=\tan (\frac{\pi }{4}+\alpha )$

$\Rightarrow \frac{1+x^2}{1-x^2}=\frac{{\tan }^2(\frac{\pi }{4}+\alpha )}{1}$

Again using componendo and dividendo, we get

$\frac{(1+x^2)-(1-x^2)}{(1+x^2)+(1-x^2)}=\frac{{\tan }^2(\frac{\pi }{4}+\alpha )-1}{{\tan }^2(\frac{\pi }{4}+\alpha )+1}$

$\Rightarrow x^2=-\cos (\frac{\pi }{2}+2\alpha )=\sin 2\alpha$