A) \[\frac{n\pi }{3}+\frac{\pi }{12}\]
B) \[\frac{n\pi }{3}+\frac{7\pi }{36}\]
C) \[n\pi +\frac{7\pi }{12}\]
D) \[n\pi +\frac{\pi }{12}\]
Correct Answer: B
Solution :
\[\frac{\tan 3\theta -1}{\tan 3\theta +1}=\sqrt{3}\] Þ \[\frac{\tan 3\theta -\tan (\pi /4)}{1+\tan 3\theta \,.\,\tan (\pi /4)}=\sqrt{3}\] Þ \[\tan \,\left( 3\theta -\frac{\pi }{4} \right)=\tan \,\frac{\pi }{3}\] Þ \[3\theta -(\pi /4)=n\pi +(\pi /3)\] Þ \[3\theta =n\pi +\frac{7\pi }{12}\]Þ \[\theta =\frac{n\pi }{3}+\frac{7\pi }{36}\].
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