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Let A,B,C,D be positive angles such that A+B+C+D=180° show that sinAsinB+sinCsinD=sin(B+C)sin(B+D)
sinA.sinB+sinC.sinD = > 2sinA.sinB+ 2sinC.sinD => cos(A-B) – cos(A+B) + cos(C-D) – cos(C+D) -2sin(A-B+A+B)/2sin(A-B-A-B)/2 + 2sin(C-D+C+D)/2sin(C-D-C-D)/2=> -2sinA RegardsArun
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