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.
You will be redirected in
3 sec
