A) \[\frac{a-b}{1+ab}\]
B) \[\frac{b}{1+ab}\]
C) \[\frac{b}{1-ab}\]
D) \[\frac{a+b}{1-ab}\]
Correct Answer: D
Solution :
Use, \[{{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}=2{{\tan }^{-1}}x\] \[\Rightarrow \] \[2{{\tan }^{-1}}a+2{{\tan }^{-1}}b=2{{\tan }^{-1}}x\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{a+b}{1-ab} \right)={{\tan }^{-1}}x\] \[\Rightarrow \] \[x=\frac{a+b}{1-ab}\]You need to login to perform this action.
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