lf A,B,C and D are the angles of a cyclic quadrilateral, then cos A + cos B + cos C + cos D is equal to

Solution:

Given that, ABCD is a cyclic quadrilateral.

$\begin{array}{ll}\text{ So, }& \mathrm{A}+\mathrm{C}={180}^{\circ }\Rightarrow \mathrm{A}={180}^{\circ }-\mathrm{C}\\ \Rightarrow & \cos \mathrm{A}=\cos \left({180}^{\circ }-\mathrm{C}\right)=-\cos \mathrm{C}\\ \Rightarrow & \cos \mathrm{A}+\cos \mathrm{C}=0---\left(\mathrm{i}\right)\\ \text{ Similarly, }& \cos \mathrm{B}+\cos \mathrm{D}=0----\left(\mathrm{ii}\right)\end{array}$

On adding Eqs. (i) and (ii), we get
cosA + cosB + cosC + cosD =0