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 

We will proceed by simplifying the numerator and denominator separately and repeatedly use some well-known formulae.

For simplifying the numerator: \begin{align} \sin 2A + \sin 2B + \sin 2C & = 2\sin(A+B)\cos(A-B) + 2\sin C \cos C \\ & = 2\sin(\pi - C) \cos(A - B) + 2\sin C \cos C \\ & = 2\sin C [\cos(A - B) + \cos(\pi - (A + B))] \\ & = 2\sin C[\cos(A - B) - \cos(A + B)] \\ & = 4\sin A \sin B \sin C\end{align}
For simplifying the denominator: \begin{align} \cos A +\cos B + \cos C - 1 & = 2\cos(\frac{A +B}{2}) \cos(\frac{A-B}{2}) - 2 \sin^2 \frac{C}{2} \\ & = 2\cos(\frac{\pi}{2} - \frac{C}{2})\cos(\frac{A-B}{2}) - 2\sin^2\frac{C}{2} \\ & = 2\sin \frac{C}{2}[\cos(\frac{A-B}{2}) - \sin \frac{C}{2}] \\ & =2\sin \frac{C}{2}[\cos(\frac{A-B}{2}) - \cos(\frac{A+B}{2})] \\ & =4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\end{align}

Can you take it from here? 
