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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..

bigfocalchord
  • 5,597

1 Answers1

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

Max Payne
  • 3,447