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Question

In a ABC, sin2Asin2B+sin2C is equal to

A
2sinAsinBcosC
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B
2sinAcosBsinC
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C
2sinAcosBcosC
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D
2sinAsinBsinC
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Solution

The correct option is B 2sinAcosBsinC
sin2Asin2B+sin2C=sin(A+B)sin(AB)+sin2C(sin(A+B)=sin(πC)=sinC)=sinC(sin(AB)+sinC)=sinC(sin(AB)+sin(A+B))=2sinAcosBsinC

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