Question

Let $ABCD$ be a circular quadrilateral then prove that $\cos A+\cos B+\cos C+\cos D=0$.

Answer 1

Since $ABCD$ is a circular quadrilateral then we have the following relations

$A+C={180}^o$......(1) and $B+D={180}^o$.......(2).

Now,

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

$=\cos A+\cos B+\cos ({180}^o-A)+\cos ({180}^o-B)$ [ Using (1) and (2)]

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

$=0$.