Express $x$ in terms of $a$ and $b$: $\sin^{-1} {\frac{2a}{1+a^2}} + \sin^{-1}{\frac{2b}{1+b^2}} = 2\tan^{-1}x$ 
Find the value of $x$ from the following equation in terms of $a$ and $b$
  $$\sin^{-1} {2a\over{1+a^2}} + \sin^{-1}{2b\over{1+b^2}} = 2\tan^{-1}x$$ 

I tried to expand the LHS using the formula $$\sin^{-1}c+\sin^{-1}d = \sin^{-1}\left(c\sqrt{1+d^2} + d\sqrt{1+c^2}\right)$$
But it didn't work out. Could anyone help me out?

Divide both sides by $2$, then using the $\arctan(x)=y \implies x=\tan(y)$ you will get that $$x=\tan\left(\frac{\arcsin\left(\frac{2a}{1+a^2}\right)+\arcsin\left(\frac{2b}{1+b^2}\right)}{2}\right)$$
