tan(3x/2)-tan(x/2)=sin(3x/2)/cos(3x/2)-sin(x/2)/cos(x/2)(通分)=[sin(3x/2)cos(x/2)-cos(3x/2)sin(x/2)]/[cos(3x/2)cos(x/2)]=sin(3x/2-x/2]/[(1/2)(cos2x+cosx)(积化和差)=2sinx/(cosx+cos2x)故原式成立....