在三角形ABC中,已知(sinA+sin+B+sinC)(sinA+sinB-sinC)=3,a

∵acosA+bcosB=ccosC
∴sinAcosA+sinBcosB=sinCcosC
∴sin2A+sin2B=sin2C=sin(2π-2A-2B)=-sin(2A+2B)
∴0=sin2A+sin2B+sin(2A+2B)
=sin2A+sin2B+sin2Acos2B+sin2Bcos2A
=sin2A(1+cos2B)+sin2B(1+cos2A)
=4sinAcosA(cosB)^2+4sinBcosB(cosA)^2
=4cosAcosBsin(A+B)
∵sin(A+B)=sin(π-C)=sinC>0,且a