(tanA+tanB)/(1-tanAtanB)=3(tanA-tanB)/(1+tanAtanB)==>2tanA-4tanAtanAtanB-4tanB+2tanAtanBtanB=0
==>(sinAcosAcosBcosB-2cosAcosAsinBcosB-2sinAsinAsinBcosB+sinAcosAsinBsinB)=0==>设X=sinAcosA,Y=sinBcosB==>(sinBsinB+cosBcosB)X-2(sinAsinA+cosAcosA)Y=0==>X/Y=1/2==>sin2A/sin2B=(2sinAcosA)/(2sinBcosB)=X/Y=1/2