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