If $A+B+C+D=\pi$ then find $\sum\cos A\cos B-\sum\sin A\sin B$ Question:
If $A+B+C+D=\pi$ then find $\sum\cos A\cos B-\sum\sin A\sin B$
My Attempt:
$$\cos((A+B)+(C+D))=-1\\\implies\cos(A+B)\cos(C+D)-\sin(A+B)\sin(C+D)=-1\\\implies(\cos A\cos B-\sin A\sin B)(\cos C\cos D-\sin C\sin D)-(\sin A\cos B+\cos A\sin B)(\sin C\cos D+\sin D\cos C)=-1\\$$
By opening the brackets, I am not getting the desired expression. How to approach this?

Using $\cos(x+y)=\cos x\cos y-\sin x\sin y$, we have
$$\cos A\cos B-\sin A\sin B=\cos(A+B),$$
$$\cos B\cos C-\sin B\sin C=\cos(B+C),$$
$$\cos C\cos D-\sin C\sin D=\cos(C+D),$$
$$\cos D\cos A-\sin D\sin A=\cos(D+A),$$
hence
$$\sum\cos A\cos B-\sum\sin A\sin B=\cos(A+B)+\cos(C+D)+\cos(B+C)+\cos(D+A).$$
Now using $\cos(\pi-x)+\cos x=0$, we have
$$\cos(A+B)+\cos(C+D)=0\qquad\text{and}\qquad\cos(B+C)+\cos(D+A)=0.$$
