inv-trig

Question: If ${{\sin }^{-1}}\left( \frac{2a}{1+{{a}^{2}}} \right)+{{\sin }^{-1}}\left( \frac{2b}{1+{{b}^{2}}} \right)=2{{\tan }^{-1}}x,$then $x=$



1) $\frac{a-b}{1+ab}$
2) $\frac{b}{1+ab}$
3) $\frac{b}{1-ab}$
4) $\frac{a+b}{1-ab}$
Solution: Explanation: No Explanation
Inverse trigonometric functions

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