A) \[n\pi \]
B) \[\frac{n\pi }{6}\]
C) \[n\pi -\frac{\pi }{4}\pm \alpha \]
D) \[\frac{n\pi }{2}\]
Correct Answer: B
Solution :
\[\tan \theta +\tan 2\theta +\tan 3\theta =\tan \theta \tan 2\theta \tan 3\theta \] \[\tan 6\theta =\frac{\tan \theta +\tan 2\theta +\tan 3\theta -\tan \theta \tan 2\theta \tan 3\theta }{1-\sum \tan \theta \tan 2\theta }\] = 0, (from the given condition) \[\Rightarrow \] \[6\theta =n\pi \Rightarrow \theta =n\pi /6\]. Trick: In such type of problems, the general value of \[\theta \] is given by\[\frac{n\pi }{\text{sum of number of }\theta }\]. So the general value of \[\theta \] is\[\frac{n\pi }{1+2+3}=\frac{n\pi }{6}\].
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