Question

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

• A. $1$
• B. $-1$
• C. $\pm 1$
• D. $0$

Answer 1

The correct option is D $0$

In cyclic quadrilateral sum of opposite angles is ${180}^{\circ }$

$A+C={180}^{\circ }$ & $B+D={180}^{\circ }$

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

$cosA+cosC=2cos\left(\frac{180}{2}\right)cos\left(\frac{A-C}{2}\right)$

$cosA+cosC=0$

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

$cosB+cosD=2cos\left(\frac{180}{2}\right)cos\left(\frac{B-D}{2}\right)$

$cosB+cosD=0$

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