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$\tan^{- 1} \frac{2 x}{1 - x^{2}} = \sin^{- 1} \frac{2 a}{1 + a^{2}} + \cos^{- 1} \frac{1 - b^{2}}{1 + b^{2}}$ সমীকরণে x = ?

\frac{a - b}{1 + a b}

\frac{a + b}{1 - a b}

\frac{2 a b}{1 + a b}

\frac{2 a b}{1 - a b}

ANS : 2

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