If $\prod_{cyclic} \tan\left(\frac{\alpha+\beta-\gamma}{4}\right)=1$ and $\sum \alpha = \pi $ Then find the value of $\sum \cos(\alpha )$

Question

If $$\prod_{cyclic} \tan\left(\frac{\alpha+\beta-\gamma}{4}\right)=1$$ and $\sum \alpha = \pi $

Then find the value of $\sum \cos(\alpha )$

Answer provided = $-1$ (Not sure)!!

Source: A grade 12 test

My efforts:

Using the first condition I was able to manipulate till $\prod_{cyclic} \tan(\alpha/2)$+ $\sum \tan(\alpha/2) = 0$ But was struck further.

My friend showed me his solution, he manipulated this same thing for two pages and got the expression required. Unfortunately this was 2 months ago and he says he has lost that paper and is too lazy to solve again.

I even tried hit and trial from here but neither an angle of 0 or $\pi/2$ worked.

My requirement: A decent solution.

Answer 1

Say we denote $$x, y, z = \tan\left(\dfrac{\alpha}{2}\right), \tan\left(\dfrac{\beta}{2}\right), \tan\left(\dfrac{\gamma}{2}\right).$$

Using the fact that $\sum_{cyc}\alpha=\pi$, we can see that $$\tan\left(\dfrac{\alpha+\beta-\gamma}{4}\right)=\tan\left(\dfrac{\pi}{4}-\dfrac{\gamma}{2}\right) = \dfrac{1-z}{1+z}$$ and so on for each term. The product equalling unity can then be simplified as $$(1+x)(1+y)(1+z)=(1-x)(1-y)(1-z)\:\:\text{or, } x+y+z+xyz=0.$$

We shall note that none of the three angles $\alpha, \beta,\gamma$ can be $\pi$ itself as that doesn't satisfy the product relation. Therefore, assuming principal solutions in $[0,\pi]$, we have $x,y,z<\infty$.

Check that $z=\tan(\gamma/2) = \cot(\alpha/2+\beta/2)=(1-xy)/(x+y)$.

We can then eliminate $z$ to get $-\dfrac{x+y}{1+xy}=\dfrac{1-xy}{x+y}$, and so $(x+y)^2+1=x^2y^2$.

Using the relation(s) $\cos(\alpha) = \dfrac{1-x^2}{1+x^2}$ (and so on...) we should be able to arrive at the simplification

$$\sum_{cyc}\dfrac{1-x^2}{1+x^2} = \underbrace{2\dfrac{1-x^2y^2}{(1+x^2)(1+y^2)}}_{\text{First two terms}}+\dfrac{x^2y^2-1-(1-xy)^2}{x^2y^2-1+(1-xy)^2} = \dfrac{1-x^2y^2}{xy(xy-1)}+\dfrac{1}{xy} = -1$$ where we have assumed that $xy\neq 1$ or $\gamma\neq 0$.

Note that if $xy=1$, we get $x=-y,\:z=0$, which has no real solutions.

So $\sum_{cyc}\cos(\alpha) = -1$.