A) \[\frac{\pi }{2}\]
B) \[\pi \]
C) \[\frac{\pi }{6}\]
D) \[\frac{\pi }{3}\]
Correct Answer: D
Solution :
\[2{{\tan }^{-1}}(\cos x)\]\[={{\tan }^{-1}}(\cos \text{e}{{\text{c}}^{2}}x)\] Þ \[{{\tan }^{-1}}\left( \frac{2\cos x}{1-{{\cos }^{2}}x} \right)={{\tan }^{-1}}\left( \frac{1}{{{\sin }^{2}}x} \right)\] \[\Rightarrow \frac{2\cos x}{{{\sin }^{2}}x}=\frac{1}{{{\sin }^{2}}x}\]Þ \[2\cos x=1\] \[\Rightarrow x=\frac{\pi }{3}\].
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