Long inverse trignometry If we are given the following equation: 
$$y =\arctan \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}$$
Then we have to prove $x^2 = \sin2y$

I tried by multiplying the eqiuation by 2 on both side  then tried to convert arctan into arcsin. 

But it becoes too long to solve  

Is there any short method.
HINT:

$$\tan y=\dfrac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}$$

Use  Weierstrass substitution
You may use componendo and divedendo.
Thoughts:- 

if $a/b=c/d$ then ($a+b)/(a-b)=(c+d)/(c-d)$ and vice-verca.

so $\frac{\tan y + 1}{\tan y - 1}= - \left(\frac{1+x^2}{1-x^2}\right)^{1/2}$

square both sides and then apply componendo &amp; dividendo again to get the desired result.