A) \[\frac{x}{1-{{x}^{2}}}\]
B) \[\frac{x}{1+{{x}^{2}}}\]
C) \[\frac{x}{\sqrt{1-{{x}^{2}}}}\]
D) \[\frac{1}{\sqrt{1-{{x}^{2}}}}\]
E) \[\frac{x}{\sqrt{{{x}^{2}}-1}}\]
Correct Answer: C
Solution :
\[\tan ({{\sin }^{-1}}x)=\tan \left( {{\tan }^{-1}}\frac{x}{\sqrt{1-{{x}^{2}}}} \right),x\in (-1,1)\] \[=\frac{x}{\sqrt{1-{{x}^{2}}}}\]
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