已知x+y=60°,其中x≥0°及y≥0°.求tanx+tany的最大值和最小值 z=tanx+tany=tanx+tan(π/3-x)=sinx/cosx+sin(π/3-x)/cos(π/3-x)=[sinxcos(π/3-x)+cosxsin(π/3-x)]/cosxcos(π/3-x)=sin(π/3)/cosxcos(π/3-x)=(...