Mathematics Trigonometry Inverse Trigonometric Functions

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

1
\(\frac{a-b}{1+ab}\)
2
\(\frac{a-b}{1-ab}\)
3
\(\frac{2ab}{a+b}\)
4
\(\frac{a+b}{1-ab}\)

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