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

#### Solution

$(\tan \theta +\tan 2\theta)-\dfrac{\tan\theta +\tan 2\theta}{1-\tan\theta \tan 2\theta}=0$
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:
$\theta =\dfrac{2m\pi}{2}=m\pi, \theta =\dfrac{n\pi}{3}, m, n\in I$.