A) \[\frac{2x}{\sqrt{1-{{x}^{2}}}}\]
B) \[\frac{2x}{1-{{x}^{2}}}\]
C) \[\frac{2x}{1+{{x}^{2}}}\]
D) None of these
Correct Answer: C
Solution :
Given that \[A={{\tan }^{-1}}x\] Now\[x=\tan A\Rightarrow \sin 2A=\frac{2\tan A}{1+{{\tan }^{2}}A}=\frac{2x}{1+{{x}^{2}}}\].
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