# Prove that : $\tan 40 + \tan 60 + \tan 80 = \tan 40 \cdot \tan 60 \cdot \tan 80$ [duplicate]

I started from Left hand side as 3^1/2 + tan 2(20) +tan 4(20). But that brought me a lot of terms to solve which ends (9 tan 20 - 48 tan^3 20 -50 tan^5 20 - 16 tan^7 20 + tan^9 20)/(1- 7 tan^2 20 + 7 tan^4 20 - tan^6 20), which is very huge to solve. If someone can help me in teaching me how to begin, that would be very greatful.

Generally, if $\alpha+\beta+\gamma=180^\circ$ then $\tan\alpha+\tan\beta+\tan\gamma=\tan\alpha\tan\beta\tan\gamma$.
That is because $$0 = \tan180^\circ = \tan(\alpha+\beta+\gamma) = \text{a certain fraction}.$$ The numerator in the fraction is $\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma$.
• Maybe I should add this: One should take the range of the tangent function to be $\mathbb R\cup\{\infty\}$, where $\infty$ is neither $+\infty$ nor $-\infty$, but is at both ends of the line, so that this is topologically a circle. That's not how it's usually done in textbooks. But that way $\tan$ is an everywhere continuous function and if $\alpha,\beta,\gamma$ are the angles of a right triangle, then the identity still holds. – Michael Hardy Sep 26 '13 at 22:13