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

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:
$$\theta =\dfrac{2m\pi}{2}=m\pi, \theta =\dfrac{n\pi}{3}, m, n\in I$$.