TITLE: Trig identity problem from Gelfand's Trigonometry
QUESTION [5 upvotes]: Having a problem with an exercise from Gelfand and Saul's Trigonometry, in the section dealing with the half-angle formulae. The exercise (7.a. on p.151) asks the reader to show that:
$$\tan(\alpha/2)\tan(\beta/2)\ +\ \tan(\beta/2)\tan(\gamma/2)\ +\ \tan(\gamma/2)\tan(\alpha/2)\ =\ 1$$ 
where $\alpha + \beta + \gamma = \pi$.
I made the cotangent substitution for one of the variables and checked it on Wolfram Alpha, which confirms the identity, but for some reason I still can't see how to actually prove it myself. I'm sure I'm missing something very obvious!

REPLY [2 votes]: Since $$\tan (x+y) = \frac{\tan x + \tan y}{1-\tan x \tan y}$$ we have $$\tan x + \tan y = \tan (x+y) [1-\tan x \tan y].$$
Therefore
$$\tan \frac{\alpha}{2}\left( \tan \frac{\beta}{2} + \tan \frac{\gamma}{2} \right) = \tan\left(\frac{\pi}{2} - \frac{\beta}{2}-\frac{\gamma}{2} \right) \tan\left(\frac{\beta}{2}+\frac{\gamma}{2}\right)\left[1-\tan\frac{\beta}{2} \tan\frac{\gamma}{2}\right] = 1-\tan\frac{\beta}{2} \tan\frac{\gamma}{2}.$$