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) \[{{\sec }^{-1}}x\]
Correct Answer: C
Solution :
\[{{\tan }^{-1}}\frac{1}{\sqrt{{{x}^{2}}-1}}={{\tan }^{-1}}\frac{1}{\sqrt{\text{cose}{{\text{c}}^{2}}\theta -1}}\] (Putting \[x=\text{cos}\text{ec}\,\,\theta )\] \[={{\tan }^{-1}}\frac{1}{\cot \theta }=\theta =\text{cose}{{\text{c}}^{-1}}x\].
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