Question

Prove that:
(i) $\tan A+\tan ({60}^{\circ }+A)+\tan ({120}^{\circ }+A)=3\tan 3A$.

Answer 1

$\tan A+\tan (60+A)+\tan (120+A)$

$=\tan A+\frac{\tan 60+\tan A}{1+\tan 60\tan A}+\frac{\tan 120+\tan A}{1-\tan 120.\tan A}$

$=7+\frac{\sqrt{3}+7}{1-\sqrt{3}7}+\frac{-\sqrt{3}+7}{1+\sqrt{3}7}$ [Let $\tan A=4$]

$=7+\frac{(\sqrt{3}+7)(1+\sqrt{3}7)+(1-\sqrt{3})(1-\sqrt{3}7)}{(1-\sqrt{3}7)(1+\sqrt{3}7)}$

$=7+\frac{(\sqrt{3}+47+\sqrt{3}{7}^2)+(-\sqrt{3}+47-\sqrt{3}{7}^2)}{1-{37}^2}$

$=7+\frac{87}{1-{37}^2}$

$=\frac{7-{37}^3+87}{1-{37}^2}=\frac{97-{37}^3}{1-{37}^2}=\frac{3(3\tan A-{\tan }^{3}A)}{1-3{\tan }^{2}A}=3\tan 3A$