Question

If $A,B,C$ and $D$ are angles of a cyclic quadrilateral prove that
$\cos A+\cos B+\cos C+\cos D=0$

Answer 1

Since the quadrilateral $ABCD$ is cyclic, we have

$A+C={180}^o;B+D={180}^o$

Hence, $\cos A=\cos ({180}^o-C)=-\cos C...\left(1\right)$ and $\cos B=\cos ({180}^o-C)=-\cos D...\left(2\right)$

Adding (1) and (2) we get

$\cos A+\cos B=-\cos C-\cos D$

or $\cos A+\cos B+\cos C+\cos D=0$