Question

If $A,B,C,D$ are the angles ore cyclic quadrilateral, find $\cos A+\cos B+\cos C+\cos D$.

• A. 4
• B. 1
• C. 0
• D. -1

Answer 1

The correct option is C 0

Let $A,B,C,D$ be angles taken in order.

Thus $A+C=180,B+D=180$

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

$2\cos \left(\frac{A+C}{2}\right).\cos \left(\frac{A-C}{2}\right)+2\cos \left(\frac{B+D}{2}\right).\cos \left(\frac{B-D}{2}\right)$

$2\cos \left(\frac{180}{2}\right).\cos k+2\cos \left(\frac{180}{2}\right).\cos P$

$=0$