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