We know that
A+B+C=πsin2A+sin2B+sin2C=4sinAsinBsinC
sinA+sinB+sinC=2sinA+B2cosA−B2+2sinC2cosC2(∵sin(A+B2)=sin(π2−C2)=cosC2)=2cosC2[cosA−B2+sinC2]=2cosC2[cosA−B2+cosA+B2](∵sin(C2)=sin(π2−A+B2)=cosA+B2)=4cosA2cosB2cosC2
Now,
sin2A+sin2B+sin2CsinA+sinB+sinC=4sinAsinBsinC4cosA2cosB2cosC2=8sinA2cosA2sinB2cosB2sinC2cosC2cosA2cosB2cosC2=8sinA2sinB2sinC2∴k=8