Question

Show that $\tan \theta +\tan ({60}^{\circ }+\theta )+\tan ({120}^{\circ }+\theta )=3\tan 3\theta$.

Answer 1

$\tan \theta +\tan ({60}^{\circ }+\theta )+\tan ({120}^{\circ }+\theta )$

using trigonometric identites

$=\tan \theta +\frac{\tan {60}^{\circ }+\tan \theta }{1-\tan {60}^{\circ }\tan \theta }+\frac{\tan {120}^{\circ }+\tan \theta }{1-\tan {120}^{\circ }\tan \theta }$

$=\tan \theta +\frac{\sqrt{3}+\tan \theta }{1-\sqrt{3}\tan \theta }+\frac{-\sqrt{3}+\tan \theta }{1+\sqrt{3}\tan \theta }$

$=\frac{\tan \theta (1-3{\tan }^2\theta )+\sqrt{3}+\tan \theta -\sqrt{3}+\tan \theta }{1-3{\tan }^2\theta }$

$=\frac{9\tan \theta -3{\tan }^3\theta }{1-3{\tan }^2\theta }=3\tan 3\theta$