Question

$\tan \theta +\tan 2\theta =\tan 3\theta$.

Answer 1

$(\tan \theta +\tan 2\theta )-\frac{\tan \theta +\tan 2\theta }{1-\tan \theta \tan 2\theta }=0$
or $(\tan \theta +\tan 2\theta )(1-\tan \theta \tan 2\theta -1)=0$
or $\tan \theta \tan 2\theta (\tan \theta +\tan 2\theta )=0$
$\tan \theta =0,\therefore \theta =n\pi ,\tan 2\theta =0$,
$\therefore 2\theta =n\pi$ or $\theta =n\pi /2$
$\tan \theta +\tan 2\theta =0$
or $\sin (\theta +2\theta )=0$
$\therefore \theta =n\pi /3$
But for odd values of n, the values of $\theta$ given by $\theta =n\pi /2$ do not satisfy the given equation as it will lead to $\infty =\infty$
Hence the required solution is:
$\theta =\frac{2m\pi }{2}=m\pi ,\theta =\frac{n\pi }{3},m,n\in I$.