如果实数x、y满足 |tanx|+|tany|>|tanx+tany|,且y∈(π, 3π 2...

正切函数定义域：{x|x≠(π/2)+kπ,k∈Z}值域：R最值：无最大值与最小值零值点：(kπ,0) 周期：kπ,k∈Z增区间:{x|(-π/2)+kπ<x<(π/2)+kπ,k∈Z}

令a=tanx 则a属于R y=f(x)=(a²-a+1)/(a²+a+1)ya²+ya+y=a²-a+1 (y-1)a²+(y+1)a+(y-1)=0 a是实数则方程有解 所以判别式大于等于0 (y+1)²-4(y-1)²>=0 (y+1+2y-2)(y+1-2y+2)>=0 (3y-1)(y-3)<=0 1/3<...