\[{{\tan }^{-1}}\frac{1}{\sqrt{{{x}^{2}}-1}}=\]

1.	\[{{\tan }^{-1}}\frac{1}{\sqrt{{{x}^{2}}-1}}=\]
A.	\[\frac{\pi }{2}+\text{cose}{{\text{c}}^{-1}}x\]
B.	\[\frac{\pi }{2}+{{\sec }^{-1}}x\]
C.	\[\text{cose}{{\text{c}}^{-1}}x\]
D.	\[{{\sec }^{-1}}x\]
Answer» D. \[{{\sec }^{-1}}x\]

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\[{{\tan }^{-1}}\frac{1}{\sqrt{{{x}^{2}}-1}}=\]

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