If ABCD is a cyclic quadrilateral then prove that sin A+sin B=sin C+sin D
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In a cyclic quadrilateral, sum of opposite angles is 180.
So, A+C=180⇒C=180−A⇒sinC=sin(180−A)=sinA
B+D=180⇒D=180−B⇒sinD=sin(180−B)=sinB
So,
sinA+sinB−sinC−sinD
=sinA+sinB−sinA−sinB
=0
Hence, A is correct.
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