Value of $\cos(\alpha+\beta)+\cos(\beta+\gamma)+\cos(\gamma+\alpha) =$ 
If $\alpha,\beta,\gamma$ be three distinct real values such that $$\displaystyle \frac{\sin \alpha+\sin \beta+\sin \gamma}{\sin(\alpha+\beta+\gamma)} = \frac{\cos \alpha+\cos \beta+\cos \gamma}{\cos(\alpha+\beta+\gamma)}=2$$
  Then value of $\cos(\alpha+\beta)+\cos(\beta+\gamma)+\cos(\gamma+\alpha) = $

$\bf{My\; Try::}$ Using 
$\sin (\alpha+\beta+\gamma) = \sin \alpha \cos \beta \cos \gamma+\cos \alpha \sin \beta \cos \gamma+\cos \alpha \cos \beta \sin \gamma-\sin \alpha \sin \beta \sin\gamma$
$\cos (\alpha+\beta+\gamma) = \cos \alpha \cos \beta \cos \gamma-\sin \alpha \sin \beta \cos \gamma-\cos \alpha \sin \beta \sin \gamma-\sin \alpha \cos \beta \sin \gamma$
But putting that values in above expression form very lengthy expression,
How can we solve it in less complex way,
Thanks

The three expressions are symmetric wrt $\alpha, \, \beta, \, \gamma$, which means that,
 if they are valid for a given triple, they shall be valid for any permutation of that triple.
Thus if $(\alpha, \, \beta, \, \gamma)$ is a solution that lies on the plane $\alpha + \beta + \gamma=s$, the permutations will also lie on that plane.
Therefore
$$
\eqalign{
  & \cos \left( {\alpha  + \beta } \right) + \cos \left( {\beta  + \gamma } \right) + \cos \left( {\alpha  + \gamma } \right) =   \cr 
  &  = \cos \left( {s - \alpha } \right) + \cos \left( {s - \beta } \right) + \cos \left( {s - \gamma } \right) =   \cr 
  &  = \cos s\;\left( {\cos \alpha  + \cos \beta  + \cos \gamma } \right) + \sin s\left( {\sin \alpha  + \sin \beta  + \sin \gamma } \right)  \cr 
  &  = \cos ^{\,2} s\;{{\left( {\cos \alpha  + \cos \beta  + \cos \gamma } \right)} \over {\cos s}}
 + \sin ^{\,2} s{{\left( {\sin \alpha  + \sin \beta  + \sin \gamma } \right)} \over {\sin s}} =   \cr 
  &  = 2\left( {\cos ^{\,2} s + \sin ^{\,2} s} \right) = 2 \cr} 
$$
which is independent of $s$, and is therefore the answer.
