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Question

In a ABC the expression sin2A+sin2B+sin2CsinA+sinB+sinC=ksinA2sinB2sinC2 , then the value of k is

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Solution

We know that
A+B+C=πsin2A+sin2B+sin2C=4sinAsinBsinC

sinA+sinB+sinC=2sinA+B2cosAB2+2sinC2cosC2(sin(A+B2)=sin(π2C2)=cosC2)=2cosC2[cosAB2+sinC2]=2cosC2[cosAB2+cosA+B2](sin(C2)=sin(π2A+B2)=cosA+B2)=4cosA2cosB2cosC2

Now,
sin2A+sin2B+sin2CsinA+sinB+sinC=4sinAsinBsinC4cosA2cosB2cosC2=8sinA2cosA2sinB2cosB2sinC2cosC2cosA2cosB2cosC2=8sinA2sinB2sinC2k=8

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