A) \[{{\sec }^{-1}}p\]
B) \[{{\cos }^{-1}}p\]
C) \[{{\tan }^{-1}}p\]
D) None of these
Correct Answer: A
Solution :
\[\tan \theta =\frac{\pm 2\sqrt{{{p}^{2}}-1}}{1+1}=\pm \sqrt{{{p}^{2}}-1}\Rightarrow \theta ={{\sec }^{-1}}p\].
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