求y=(tanx+2)/(secx-1)的值域

1个回答

  • 上下同乘以cosx

    y=(sinx+2cosx)/(1-cosx)

    利用sinx=2sin(x/2)cos(x/2)

    cosx=sin(x/2)^2-cos(x/2)^2

    有y=(2sin(x/2)cos(x/2)+2(sin(x/2)^2-cos(x/2)^2))/(sin(x/2)^2+cos(x/2)^2-(sin(x/2)^2-cos(x/2)^2))

    化简y=tan(x/2)^2+tan(x/2)-1

    令tan(x/2)=t

    y=t^2+t-1

    t属于全体实数

    所以t=-1/2时y取最小值-5/4

    所以y>=-5/4

    附图一张