在△ABC中,角A,B,C的所对边长分别为a,b,c,a=2根号3,tan((A+B)/2)+tan(C/2)=4,sinB*sinC=cos^2(A/2)

tan[（A+B）/2]+tanC/2=4,tan(A+B)=tan(π-C）
tan[（A+B）/2]+tanC/2=tan[π/2-C/2]+tanC/2=cot(C/2)+tan(C/2)
=cos(C/2)/sin(C/2)+sin(C/2)/cos(C/2)=[sin(C/2)^2+cos(C/2)^2]/(sinC/2)(cosC/2)=2/sinC=4,C=π/6或5π/6
cos(A/2)=cos[π-(B+C)]/2=sin[(B+C)/2]
cos(A/2)^2=sin[(B+C)/2]^2=[1-cos(B+C)}/2=sinBsinC,1-cosBcosC+sinBsinC=2sinBsinC
cos(B-C)=1,B-C=0(π不符合题意）
B=C=π/6,5π/6应舍去（不能有二个钝角）,