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