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?

Let $\arctan a=u,a=\tan u$
If $-\dfrac\pi2\le2u\le\dfrac\pi2\iff -1\le a\le1$ $$P=\arcsin\dfrac{2a}{1+a^2}=2\arctan a$$
If $2u>\dfrac\pi2,P=\pi-2\arctan a$
If $2u<-\dfrac\pi2,P=-\pi-2\arctan a$
Now use my answer in showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$
