$\tan \theta+\tan 2\theta+\sqrt 3\tan \theta\tan 2\theta=\sqrt 3$
$\tan \theta+\tan 2\theta=\sqrt 3(1-\tan \theta\tan 2\theta)$
$\large\frac{\tan \theta+\tan 2\theta}{1-\tan \theta\tan 2\theta}$$=\sqrt 3$
$\tan 3\theta =\sqrt 3=\tan \large\frac{\pi}{3}$
$3\theta=n\pi+\large\frac{\pi}{3}$
$\theta=\large\frac{n\pi}{3}+\frac{\pi}{9}$
Hence the given statement is true.