Question

If $ta{n}^{-1}\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}=\alpha$, then $x^2$ is equal to

• A. $sin\alpha$
• B. $cos2\alpha$
• C. $cos\alpha$
• D. $sin2\alpha$

Answer 1

The correct option is D $sin2\alpha$

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

Apply componendo and dividendo

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

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

$\frac{1+\sin 2\alpha }{1-\sin 2\alpha }=\frac{1+x^2}{1-x^2}$

Apply componendo and dividendo

$\frac{2x^2}{2}=\frac{2\sin 2\alpha }{2}$

Hence

$x^2=\sin 2\alpha$