# Prove $\tan (\alpha)+ \tan(\alpha + 60°) + \tan(\alpha + 120°) = 3\tan(3\alpha)$

Question: Prove $\tan (\alpha)$ $+$ $\tan(\alpha + 60°)$ $+$ $\tan(\alpha + 120°) = 3\tan(3\alpha)$

What I have attempted (working from the lhs) :

$$\tan (\alpha) + \tan(\alpha + 60°) + \tan(\alpha + 120°)$$

$$\tan (\alpha) + \frac{\tan(\alpha) + \tan(60°)}{1-\tan(\alpha)\tan(60°)} + \frac{\tan(\alpha) + \tan(120°)}{1-\tan(\alpha)\tan(120°)}$$

$tan (60°) = \sqrt{3}$ and $tan(120°) =-\sqrt{3}$ so

$$\tan (\alpha) + \frac{\tan(\alpha) +\sqrt{3}}{1-\sqrt{3}\tan(\alpha)} + \frac{\tan(\alpha) - \sqrt{3}}{1+\sqrt{3}\tan(\alpha)}$$

$$\tan (\alpha) + \frac{(\tan(\alpha) +\sqrt{3})(1+\sqrt{3}\tan(\alpha) )+(\tan(\alpha) - \sqrt{3})(1-\sqrt{3}\tan(\alpha))}{(1+\sqrt{3}\tan(\alpha))(1-\sqrt{3}\tan(\alpha))}$$

$$\tan (\alpha) + \frac{2\tan(\alpha) + 6\tan(\alpha)}{1-3\tan^2(\alpha)}$$

$$\frac {\tan (\alpha)(1-3\tan^2(\alpha)) + 8\tan(\alpha) }{1-3\tan^2(\alpha)}$$

$$\frac {9\tan(\alpha) - 3\tan^3(\alpha)}{1-3\tan^2(\alpha)}$$

Now I am stuck..

• $\tan(\alpha +\alpha +\alpha)=\cdots$ – Achaire Feb 16 '16 at 9:55
• – lab bhattacharjee Feb 16 '16 at 10:18

Hint: $$\tan(3a) = \frac{3 \tan a-\tan^3 a}{1-3\tan^2 a}$$ So, take 3 common from your last result: $$\frac{9 \tan a-3\tan^3 a}{1-3\tan^2 a} = 3\tan (3a)$$