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Question

Prove that
sin2Asin2BsinAcosAsinBcosB=tan(A+B)

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Solution

Let us consider the problem
sin2Asin2BsinAcosAsinBcosB=1cos2A21cos2B22sinAcosA22sinBcosB2
=1cos2A1+cos2Bsin2Asin2B

=(cos2Acos2B)sin2Asin2B

=(2sin(2A+2B2)sin(2A2B2))2sin(2A2B2)cos(2A+2B2)
=sin(A+B)cos(A+B)=tan(A+B)

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