# Trig identity problem from Gelfand's Trigonometry

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!

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We have $\dfrac\alpha2 + \dfrac\beta2 + \dfrac\gamma2 =\dfrac\pi2$.
$\cot (\dfrac\alpha2 + \dfrac\beta2 + \dfrac\gamma2) = \large\dfrac{1-\tan\frac\alpha2\tan\frac\beta2 - \tan\frac\beta2\tan\frac\gamma2 - \tan\frac\gamma2\tan\frac\alpha2}{\tan\frac\alpha2+\tan\frac\beta2+\tan\frac\gamma2-\tan\frac\alpha2\tan\frac\beta2\tan\frac\gamma2}=0$ since $\cot\dfrac\pi2=0$.
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 $$\tan \frac{\alpha}{2}\left( \tan \frac{\beta}{2} + \tan \frac{\gamma}{2} \right) = \tan\left(\frac{\pi}{2} - \frac{\beta}{2}-\frac{\gamma}{2} \right) \tan\left(\frac{\beta}{2}+\frac{\gamma}{2}\right)\left[1-\tan\frac{\beta}{2} \tan\frac{\gamma}{2}\right] = 1-\tan\frac{\beta}{2} \tan\frac{\gamma}{2}.$$