Question

If ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=\alpha$, then $x^2$ is equal to :

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

Answer 1

The correct option is A $\sin 2\alpha$

We have,

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

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

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

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

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

Let $x^2=\sin \theta$ ----------------------(i)

$\Rightarrow \frac{1+\sqrt{(1-\sin {}^2\theta )}}{\sin \theta }=\tan \alpha$

$\Rightarrow \frac{1+\cos \theta }{\sin \theta }=\tan \alpha$

Now, using identities :

$\Rightarrow \cos 2\theta =2{cos}^2\theta -1$

$\Rightarrow \sin 2\theta =2\sin \theta \cos \theta$

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

$\Rightarrow \cot \frac{\theta }{2}=\tan \alpha$

$\Rightarrow \frac{\pi }{2}-\frac{\theta }{2}=\alpha$

$\theta =\pi -2\alpha$

Using (i),

$x^2=\sin \theta$

$\Rightarrow x^2=\sin (\pi -2\alpha )$

$\Rightarrow x^2=\sin \left(2\alpha \right)$