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.