<s>[INST] Solve for $x$ from an equation containing inverse trigonometric functions 
How to solve the following for $x$?
  $$
  \sin^{-1}\left(\frac{2a}{1+a^{2}}\right)+
  \sin^{-1}\left(\frac{2b}{1+b^{2}}\right)= 
  2 \tan^{-1}(x )
$$ 

What conditions apply?
 [/INST] Hint: Let $\theta = \sin^{-1}\left(\dfrac{2a}{1+a^2}\right) \Rightarrow \sin \theta = \dfrac{2a}{1+a^2}$, and similarly $\sin \beta = \dfrac{2b}{1+b^2}$. Thus:
$\tan^{-1}x = \dfrac{\theta}{2}+\dfrac{\beta}{2} \Rightarrow x = \tan\left(\dfrac{\theta}{2}+\dfrac{\beta}{2}\right)$. You can compute $\tan\left(\dfrac{\theta}{2}\right)$, and $\tan\left(\dfrac{\beta}{2}\right)$ using the above equations, then find $x$.
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