(b) $\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 $\frac{{n\pi }}{{{\rm{sum \,of\, number\, of\, }}\theta }}$.

So the general value of $\theta $ is $\frac{{n\pi }}{{1 + 2 + 3}} = \frac{{n\pi }}{6}$.