If \(\rm \sin^{-1} \left( \frac {2a}{1+a^2} \right) + \sin^{-1} \left( \frac {2b}{1 + b^2} \right) = 2 \tan^{-1} x\), then x is equal to
1
\(\rm \frac {a - b}{1 + ab}\)
2
\(\rm \frac {a - b}{1 - ab}\)
3
\(\rm \frac {2a b}{a + b}\)
4
\(\rm \frac {a + b}{1 - ab}\)