A) \[\frac{x}{\sqrt{1+{{x}^{2}}}}\]
B) \[\frac{-x}{\sqrt{1+{{x}^{2}}}}\]
C) \[\frac{x}{\sqrt{1-{{x}^{2}}}}\]
D) None of these
Correct Answer: A
Solution :
\[y=\sec ({{\tan }^{-1}}x)\]\[\therefore \theta =\frac{1}{2}{{\cos }^{-1}}x\] \[=\frac{x}{1+{{x}^{2}}}\,.\,\sqrt{1+{{x}^{2}}}=\frac{x}{\sqrt{1+{{x}^{2}}}}\], \[({{\tan }^{-1}}x={{\sec }^{-1}}\sqrt{1+{{x}^{2}}})\].You need to login to perform this action.
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