已知直线L：xcosθ+ysinθ-1=0,若θ∈（π/2,π）则直线L的倾斜角是多少?

k=-cosθ/sinθ= [-sin(π/2-θ)]/[cos(π/2-θ) =[sin(θ-π/2)]/cos(θ-π/2)=tan(θ-π/2) 而θ-π/2∈（π/2,π）由倾斜角的定义可知：直线倾斜角为θ-π/2