Question

If A, B, C, D are angles of a cyclic quadrilateral, prove that $cosA+cosB+cosC+cosD=0$.

Answer 1

We know that the opposite angles of a cyclic quadrilateral are supplementary
i.e., $A+C=\pi$ and $B+D=\pi$
$\therefore A=\pi -C$ and $B=\pi -D$
$\Rightarrow cosA=cos(\pi -C)=-cosC$
and $cosB=cos(\pi -D)=-cosD$
$\therefore cosA+cosB+cosC+cosD$
$=-cosC-cosD+cosC+cosD=0$