Question

If ABCD is a cyclic quadrilateral then find the value of: $\cos A\cos B-\cos C\cos D$?

Answer 1

Given ABCD is a cyclic quadrilateral
$A+C={180}^{\circ }$ and $B+D={180}^{\circ }.......\left(1\right)$
$-\cos C\cos D=\cos A\cos B-\cos ({180}^{\circ }-A)$
$\cos ({180}^{\circ }-B)$
$=\cos A-\cos B-\cos A(-\cos A)(-\cos B)$
$=\cos A\cos B-\cos A\cos B=0$