Question

If $A,B,C,D$ are the angles of a cyclic quadrilateral, show that $\cos A+\cos B+\cos C+\cos D=0$.

Answer 1

Prove the statement.

Given: Quadrilateral $ABCD$ is cyclic in nature,

Thus, $A+C=180°$

$B+D=180°$

Hence, $\cos A=\cos (180°-C)$ $\left\{\cos \left(180°-\theta \right)=\cos \theta \right\}$

$\Rightarrow$ $\cos A=-\cos C$………………… $\left(1\right)$

$\cos B=\cos (180°-D)$ $\left\{\cos \left(180°-\theta \right)=\cos \theta \right\}$

$\Rightarrow$ $\cos B=-\cos D$………………… $\left(2\right)$

Add equation $1$and equation $2$.

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

$\Rightarrow \boxed{\cos A+\cos B+\cos C+\cos D=0}$

Hence proved that $\cos A+\cos B+\cos C+\cos D=0$.