A) \[\pi /12\]
B) \[\pi /6\]
C) \[\pi /4\]
D) \[\pi /3\]
Correct Answer: B
Solution :
\[\tan 2\theta .tan\theta =1\] \[\Rightarrow \,\,\,\frac{2\tan \theta }{1-{{\tan }^{2}}\theta }.\tan \theta =1\] \[\Rightarrow \,\,2{{\tan }^{2}}\theta =1-{{\tan }^{2}}\theta \Rightarrow 3{{\tan }^{2}}\theta =1\] \[\Rightarrow \,\,\,{{\tan }^{2}}\theta =\frac{1}{3}={{\left( \frac{1}{\sqrt{3}} \right)}^{2}}\] \[\Rightarrow \,\,\,{{\tan }^{2}}\theta ={{\tan }^{2}}(30{}^\circ )={{\tan }^{2}}\left( \frac{\pi }{6} \right)\] \[\left[ \because \,\,\,\theta =n\pi \pm \frac{\pi }{6} \right];\] \[\therefore \,\,\theta =\frac{\pi }{6}\]
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