A) \[\frac{3\pi }{4}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{3}\]
D) None of these
Correct Answer: B
Solution :
We have 2\[{{\tan }^{-1}}(\cos x)={{\tan }^{-1}}(2\cos \text{ec}x)\] Þ \[{{\tan }^{-1}}\left( \frac{2\cos x}{1-{{\cos }^{2}}x} \right)\]= \[{{\tan }^{-1}}(2\,\text{cosec }x)\] \[\frac{2\cos x}{{{\sin }^{2}}x}=2\text{cosec}\,x\Rightarrow 2\cos x=2\sin x\] or \[\sin x=\cos x\] \[\Rightarrow x=\frac{\pi }{4}\].
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