Question

Prove that $\tan \theta +\tan ({60}^o+\theta )+\tan ({120}^o+\theta )=3\tan 3\theta$

Answer 1

$\tan \theta +\tan ({60}^o+\theta )+\tan ({120}^o+\theta )=3\tan 3\theta LHS=\tan \theta +\tan ({60}^o+\theta )+\tan ({120}^o+\theta )=\tan \theta +\frac{(\tan {60}^o+\tan \theta )}{({1-\tan 60}^o.\tan \theta )}+\frac{({\tan 120}^o+\tan \theta )}{({1-\tan 120}^o.\tan \theta )}=\tan \theta +\frac{[\sqrt{3}+\tan \theta ]}{[1-\sqrt{3}\tan \theta ]}+\frac{[\tan \theta -\sqrt{3}]}{[1+\sqrt{3}\tan \theta ]}=\frac{\tan \theta +[\sqrt{3}+3\tan \theta +\tan \theta +\sqrt{3}{\tan }^2\theta +\tan \theta -\sqrt{3}-\sqrt{3}{\tan }^2\theta +3tan\theta ]}{[1-3{\tan }^2\theta ]}=tan\theta +\frac{\left[8\tan \theta \right]}{[1-3{\tan }^2\theta ]}=\frac{[9\tan \theta -3{\tan }^3\theta ]}{[1-3{\tan }^2\theta ]}=3\tan 3\theta =RHS$