\[\tan ({{\cos }^{-1}}x)\] is equal to [IIT 1993]

1.	\[\tan ({{\cos }^{-1}}x)\] is equal to [IIT 1993]
A.	\[\frac{\sqrt{1-{{x}^{2}}}}{x}\]
B.	\[\frac{x}{1+{{x}^{2}}}\]
C.	\[\frac{\sqrt{1+{{x}^{2}}}}{x}\]
D.	\[\sqrt{1-{{x}^{2}}}\]
Answer» B. \[\frac{x}{1+{{x}^{2}}}\]

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\[\tan ({{\cos }^{-1}}x)\] is equal to [IIT 1993]

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