Prove the triangle identity

Hi everybody, I need your help again. I have to prove the following identity:

(b-2*cos(c)*a)/(a*sin(c))+(c-2*cos(a)*b)/(b*sin(a))+(a-2*cos(b)*c)/(c*sin(b))=0

For every triangle (a,b,c represents length of sides and opposite angles)

I know I should use laws of sine and cosine but I couldn't complete it.

Also sorry for my post, I'm on my tablet so I didn't use equation editor.

$\displaystyle (b-2*a*cos(gama))/(a*sin(gama))+(c-2*b*cos(alpha))/(b*sin(alpha))+(a-2*cos(beta))/(c*sin(beta))$

Alpha - opposite of a

Beta - opposite of b

Gama - opposite of c

Re: Prove the triangle identity

It would help if you would distinguish between "A", the measure of an angle, and "a", the length of the side opposite that angle.