Question

$A,B,C,D$ are the angles of a cyclic quadrilateral $ABCD$, what is the value of $\cos A+\cos B+\cos C+\cos D$?

Answer 1

Because the opposite angles of a cyclic quadrilateral are supplementary
$\therefore A+B={180}^0$
and $\therefore B+D={180}^0$
i.e.$C={180}^0-A$
and $D={180}^0-B$
$\therefore \cos A+\cos B+\cos C+\cos D=\cos A+\cos B+\cos ({180}^0-A)+\cos ({180}^0-B)=\cos A+\cos B-\cos A-\cos B=0$