Question

If A, B, C & D are angles of cyclic quadrilateral then prove that
COSA+cos B+ COSC + cos D = 0​

Answer 1

Question:

If A , B , C, D are the angles of a cyclic Quadrilateral taken in order, prove that CosA+CosB+CosC+CosD=0.

Solution:

A,B,C,D are angles of a cyclic quadrilateral .

Therefore,

$A + C = {180}^{°} \\ \\ = > C = {180}^{°} - A$

And

$B + D = {180}^{°} \\ \\ = > D = {180}^{°} - B$

L.H.S

$= > cosA + cosB + cosC + cosD\\ \\ = > cosA + cos B + cos( {180}^{°} - A ) + cos {180}^{°} - B \\ \\ = > cosA + cosB - cos A - cosB \\ \\ = > 0$

= R.H.S

Hence, L.H.S=R.H.S