If $A+B+C=180$, then $\frac{\sin2A+\sin2B+\sin2C}{\cos A+\cos B+\cos C-1}=8\cos\frac A2 \cos\frac B2 \cos\frac C2$ Then $2A+2B+2C =360$
So $$\sin 2C=-\sin(2A+2B)$$
Putting that in the equation 
$$\frac{2\sin(A+B)\sin(A-B)-2\sin(A+B)\cos(A+B)}{\cos A+\cos B-\cos(A+B)+1}$$
$$\frac{2\sin(A+B)[\sin(A-B)-\cos(A+B)]}{\cos A+\cos B-\cos(A+B)+1}$$
I don’t know how to proceed. Please help me continue 

In a triangle, $$R^2(\sin(2A)+\sin(2B)+\sin(2C))=2\Delta$$
is a consequence of $[ABC]=[OAB]+[OBC]+[OCA]$ with $O$ being the circumcenter.
Carnot's theorem gives $R(\cos A+\cos B+\cos C-1) = r$, hence by $abc=4R\Delta$ the claim is equivalent to
$$ (a+b+c)\Delta = 2abc\cos\tfrac{A}{2}\cos\tfrac{B}{2}\cos\tfrac{C}{2}$$
which is simple to prove by squaring both sides, then exploiting Heron's formula and 
$$ \cos^2\tfrac{A}{2}=\tfrac{1}{2}\left(1+\cos A\right)=\frac{1}{2}\left(\frac{b^2+c^2-a^2}{2bc}+1\right)=\frac{(a+b+c)(-a+b+c)}{4bc}. $$
