Question

If A, B, C, D are angles of a cyclic quadrilateral , find the value of $cosA+cosB+cosC+cosD$.

• A. $0$
• B. $1$
• C. $2$
• D. $-1$

Answer 1

The correct option is A $0$

A cyclic quadrilateral is a quadrilateral whose all vertices lie on a single circle.

And in a cyclic quadrilateral, the sum of opposite angle is ${180}^0$

i.e. $\angle A+\angle C={180}^0$ and $\angle B+\angle D={180}^0$

$\therefore$ $cosA+cosC=2\cos \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)$

$cosA+cosC=2\cos \left(\frac{{180}^0}{2}\right).\cos \left(\frac{A-C}{2}\right)=2\cos {90}^0.\cos \left(\frac{A-C}{2}\right)=0(\because {\cos 90}^0=0)$

$cosB+cosD=2\cos \left(\frac{B+D}{2}\right).\cos \left(\frac{B-D}{2}\right)$

$=2\cos \left(\frac{{180}^0}{2}\right).\cos \left(\frac{B-D}{2}\right)$

$=2\cos {90}^0.\cos \left(\frac{B-D}{2}\right)$

$=0$ $(\because {\cos 90}^0=0)$

$\therefore$ $cosA+cosC+cosB+cosD=0$

Hence proved.