If $A+B+C=180^{\circ}$, then show that: $\cos B=\sin A\sin C-\cos A\cos C$ Here is the question :

If $A+B+C = 180^{\circ}$, then show that: $\cos B = \sin A \sin C - \cos A \cos C$.

EDIT : Here is my reviewed working :
$$
\cos B=-\cos (A+C)
$$
Since $$\space A+B+C = 180^\circ, \space B =180^\circ-(A+C)$$
And 
$$\begin{align}
-\cos B &=\cos (180+B) \\
-\cos B &=\cos(180+(180-(A+C)) \\
-\cos B &=\cos(360-(A+C)) \\
-\cos B &=\cos(A+C) \\
-\cos B &=\cos A \cos C - \sin A \sin C \\
\cos B &= \sin A \sin C - \cos A \cos C \\
\cos B &= -\cos (A+C)
\end{align}$$
Can someone confirm that my working is correct?
Thanks!

Your first step $\cos(B) = \cos(A+C)$ is incorrect. It should be $\cos(B) = -\cos(A+C)$.
