Question

$\tan \theta +\tan 4\theta +\tan 7\theta =\tan \theta \tan 4\theta \tan 7\theta$.

Answer 1

We have, $\tan \theta +\tan 4\theta +\tan 7\theta =\tan \theta \tan 4\theta \tan 7\theta$

$\tan \theta +\tan 4\theta =-\tan 7\theta (1-\tan \theta \tan 4\theta )$

$\frac{\tan \theta +\tan 4\theta }{1-\tan \theta \tan 4\theta }=-\tan 7\theta$

$\tan (\theta +4\theta )=-\tan 7\theta$ ----- Since, $\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan (A+B)$

$\tan 5\theta =\tan (-7\theta )$

$\therefore 5\theta =(-7\theta )$

$12\theta =0$ or $12\theta =n\pi$

$\therefore \theta =\frac{n\pi }{12}$, $\forall n$ as $I$