Question

if A,B,C,D are the angles of a cyclic quadrilateral, the p.t. i)cosA+cosB+cosC+cosD=0 ii)sinA-sinC=sinD-sinB.​

Answer 1

Step-by-step explanation:

Since the quadrilateral ABCD is cyclic, we have

A+C=180o

;B+D=180o

Hence, cosA=cos(180o

−C)=−cosC...(1) and cosB=cos(180o

−C)=−cosD...(2)

Adding (1) and (2) we get

cosA+cosB=−cosC−cosD

or cosA+cosB+cosC+cosD=0