Question

cosA +cosB +cosC +cosD =0​

Answer 1

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

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