Question

If $ABCD$ is a cyclic quadrilateral, then the value of $\cos A+\cos B+\cos C+\cos D$ is

• A. $0$
• B. $-1$
• C. $1$
• D. Cannot be found

Answer 1

The correct option is A $0$

Sum of the opposite angles of a cyclic quadrilateral is $\pi$
In the cyclic quadrilateral $ABCD$
$A+C=B+D=\pi$ ...$\left(i\right)$
$\cos A+\cos B+\cos C+\cos D$
$=\cos A+\cos C+\cos B+\cos D$
$=2\cos \left(\frac{A+C}{2}\right)\left(\frac{A-C}{2}\right)+2\cos \left(\frac{B+D}{2}\right)\left(\frac{B-D}{2}\right)$
$=2\cos \left(\frac{\pi }{2}\right)\left(\frac{A-C}{2}\right)+2\cos \left(\frac{\pi }{2}\right)\left(\frac{B-D}{2}\right)$ ...from $\left(i\right)$
$=0$
Hence answer is $A$