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