Prove that $1 + \cos\alpha + \cos\beta + \cos\gamma = 0$ If $\alpha + \beta + \gamma = \pi $ and $\tan(\frac{-\alpha + \beta + \gamma}4)\tan(\frac{\alpha - \beta + \gamma}4)\tan(\frac{\alpha + \beta - \gamma}4) = 1$
Then prove that:

$1 + \cos\alpha + \cos\beta + \cos\gamma = 0$.

I have no idea how to go about this. 
Please help.
 A: As $\displaystyle \frac{-\alpha+\beta+\gamma}4=\frac{-\alpha+\pi-\alpha}4=\frac\pi4-\frac\alpha2$
$\displaystyle \tan\frac{-\alpha+\beta+\gamma}4=\tan\left(\frac\pi4-\frac\alpha2\right)=\frac{1-\tan\frac\alpha2}{1+\tan\frac\alpha2}=\frac{\cos\frac\alpha2-\sin\frac\alpha2}{\cos\frac\alpha2+\sin\frac\alpha2}$
$\displaystyle\implies \tan^2\left(\frac{-\alpha+\beta+\gamma}4\right)=\left(\frac{\cos\frac\alpha2-\sin\frac\alpha2}{\cos\frac\alpha2+\sin\frac\alpha2}\right)^2=\frac{1-\sin\alpha}{1+\sin\alpha}$
$\displaystyle\implies \prod\left(\frac{1-\sin\alpha}{1+\sin\alpha}\right)=1$
$\displaystyle\implies\sum\sin\alpha+\prod\sin\alpha=0 $
Now set $\displaystyle\sin\alpha=2\sin\frac{\alpha}2\cos\frac{\alpha}2$ in $\displaystyle\prod\sin\alpha$ 
and  use this for $\sum\sin\alpha$
Cancelling out $\displaystyle\prod\cos\frac{\alpha}2,$ (assuming $\displaystyle\prod\cos\frac{\alpha}2\ne0$ )
we get $\displaystyle4\prod \sin\frac{\alpha}2=-2$
Finally, using this, $\displaystyle\sum \cos A=1+4\prod\sin\frac{\alpha}2$
A: A little Generalization :
Let $\alpha+\beta+\gamma=4C$ 
and $\displaystyle\tan\left(\frac{-\alpha + \beta + \gamma}4\right)\tan\left(\frac{\alpha - \beta + \gamma}4\right)\tan\left(\frac{\alpha + \beta - \gamma}4\right) = \cot C$
As $\displaystyle \frac{-\alpha+\beta+\gamma}4=\frac{-\alpha+4C-\alpha}4=C-\frac\alpha2,$
$\displaystyle \tan\frac{-\alpha+\beta+\gamma}4=\tan\left(C-\frac\alpha2\right)=\frac{\sin\left(C-\frac\alpha2\right)}{\cos\left(C-\frac\alpha2\right)}$
So, the problem reduces to $\displaystyle\frac{\sin\left(C-\frac\alpha2\right)\sin\left(C-\frac\beta2\right)\sin\left(C-\frac\gamma2\right)}{\cos\left(C-\frac\alpha2\right)\cos\left(C-\frac\beta2\right)\cos\left(C-\frac\gamma2\right)}=\cot C$
$\displaystyle\implies\frac{\sin\left(C-\frac\alpha2\right)\sin\left(C-\frac\beta2\right)}{\cos\left(C-\frac\alpha2\right)\cos\left(C-\frac\beta2\right)}
=\frac{\cos\left(C-\frac\gamma2\right)\cos C}{\sin\left(C-\frac\gamma2\right)\sin C}$
Applying $\displaystyle2\sin A\sin B,2\cos A\cos B$ formula,
$\displaystyle\implies\frac{\cos\frac{\beta-\alpha}2-\cos\frac{4C-\beta-\alpha}2}{\cos\frac{\beta-\alpha}2+\cos\frac{4C-\beta-\alpha}2}
=\frac{\cos\left(-\frac\gamma2\right)+\cos\frac{4C-\gamma}2}{\cos\left(-\frac\gamma2\right)-\cos\frac{4C-\gamma}2}$
Now as $\displaystyle\alpha+\beta+\gamma=4C,$ this becomes
$\displaystyle\implies\frac{\cos\frac{\beta-\alpha}2-\cos\frac{\gamma}2}{\cos\frac{\beta-\alpha}2+\cos\frac{\gamma}2}
=\frac{\cos\frac\gamma2+\cos\frac{\alpha+\beta}2}{\cos\frac\gamma2-\cos\frac{\alpha+\beta}2}$
Applying Componendo and dividendo,
$\displaystyle\implies\frac{\cos\frac{\beta-\alpha}2}{-\cos\frac{\gamma}2}
=\frac{\cos\frac{\alpha+\beta}2}{\cos\frac{\alpha+\beta}2}$
$\displaystyle\implies\cos\frac{\beta-\alpha}2\cos\frac{\beta+\alpha}2=-\cos^2\frac{\gamma}2 $
Applying $2\cos A\cos B,\cos2x=2\cos^2x-1,$
$\displaystyle\frac{\cos\alpha+\cos\beta}2=-\frac{1+\cos\gamma}2$
Here $\displaystyle C=\frac\pi4\implies \alpha+\beta+\gamma=4C=\pi$
If $\displaystyle C=-\frac\pi4\implies \alpha+\beta+\gamma=4C=-\pi\equiv\pi\pmod{2\pi}$
So, $\displaystyle\tan\left(\frac{-\alpha + \beta + \gamma}4\right)\tan\left(\frac{\alpha - \beta + \gamma}4\right)\tan\left(\frac{\alpha + \beta - \gamma}4\right)=-1$ and $\alpha+\beta+\gamma=-\pi$ or $\pi$ will satisfy the required identity  
A: Let's introduce the shorthand notations
$$
\begin{align}
a &= \frac{-\alpha+\beta+\gamma}{4} = \frac{\pi}{4} - \frac{\alpha}{2},\\
b &= \frac{\alpha-\beta+\gamma}{4} = \frac{\pi}{4} - \frac{\beta}{2},\\
c &= \frac{\alpha+\beta-\gamma}{4} = \frac{\pi}{4} - \frac{\gamma}{2},
\end{align}
$$
then
$$
a+b = \frac{\gamma}{2},\quad a+c = \frac{\beta}{2},\quad b+c = \frac{\alpha}{2},
$$
$$
a+b+c = \frac{\pi}{4},
$$
$$
\sin a\sin b \sin c = \cos a\cos b\cos c.\tag{1}
$$
Now
$$
\cos(a+b+c) = \cos(a+b)\cos(c) - \sin(a+b)\sin c= \frac{\sqrt{2}}{2},
$$
so that
$$
\cos(a+b)\cos(c) - \sin a\sin c\cos b - \sin b\sin c\cos a = \frac{\sqrt{2}}{2}
$$
add $(2\cos a\cos b\cos c)$ to both sides, and we get
$$
\cos(a+b)\cos(c) + \cos(a+c)\cos b+ \cos(b+c)\cos(a)= \frac{\sqrt{2}}{2} + 2\cos a\cos b\cos c,
$$
which reduces to
$$
\begin{multline}
\cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) +
\cos\left(\frac{\beta}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\beta}{2}\right) +\\
\cos\left(\frac{\gamma}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\gamma}{2}\right) = \frac{\sqrt{2}}{2} + 2\cos a\cos b\cos c. \tag{2}
\end{multline}
$$
On the other hand,
$$
\sin(a+b+c) = \sin(a+b)\cos c + \cos(a+b)\sin c = \frac{\sqrt{2}}{2},
$$
so that, 
$$
\sin(a+b)\cos c + \cos a\cos b\sin c = \frac{\sqrt{2}}{2} + \sin a\sin b\sin c.\tag{3}
$$
Similarly, if we take $\sin(a+b+c) = \sin((a+c)+b)$, we get
$$
\sin(a+c)\cos b + \cos a\cos c\sin b = \frac{\sqrt{2}}{2} + \sin a\sin b\sin c.\tag{4}
$$
If we add eqs. $(3)$ and $(4)$, we get
$$
\sin(a+b)\cos c + \sin(a+c)\cos b + \sin(b+c)\cos a = \sqrt{2} + 2\sin a\sin b\sin c,
$$
which reduces to (using also $(1)$)
$$
\begin{multline}
\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) +
\sin\left(\frac{\beta}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\beta}{2}\right) +\\
\sin\left(\frac{\gamma}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\gamma}{2}\right) = \sqrt{2} + 2\cos a\cos b\cos c. \tag{5}
\end{multline}
$$
Finally, we subtract eq $(5)$ from eq $(2)$:
$$
\begin{multline}
\left(\cos\left(\frac{\alpha}{2}\right)-\sin\left(\frac{\alpha}{2}\right)\right)
\cos\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) + \ldots = - \frac{\sqrt{2}}{2},
\end{multline}
$$
which is
$$
\begin{multline}
\frac{\sqrt{2}}{2}\left(\cos\left(\frac{\alpha}{2}\right)-\sin\left(\frac{\alpha}{2}\right)\right)\left(\cos\left(\frac{\alpha}{2}\right)+\sin\left(\frac{\alpha}{2}\right)\right) + \ldots + \frac{\sqrt{2}}{2} =0,
\end{multline}
$$
so that finally
$$
\cos\alpha +\cos\beta+\cos\gamma + 1 = 0.
$$
