θ∈(-π/2,0),则直线xcosθ+ysinθ+1=0的倾斜角为A.-θ B.π/2+θ C.π+θ D.π/2-θ

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θ∈(-π/2,0),则直线xcosθ+ysinθ+1=0的倾斜角为A.-θ B.π/2+θ C.π+θ D.π/2-θ
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θ∈(-π/2,0),则直线xcosθ+ysinθ+1=0的倾斜角为A.-θ B.π/2+θ C.π+θ D.π/2-θ
θ∈(-π/2,0),则直线xcosθ+ysinθ+1=0的倾斜角为
A.-θ B.π/2+θ C.π+θ D.π/2-θ

θ∈(-π/2,0),则直线xcosθ+ysinθ+1=0的倾斜角为A.-θ B.π/2+θ C.π+θ D.π/2-θ
k=-cosθ/sinθ=-cotθ
tant=cot(π/2+θ)
答案为B

化为y=-xtanθ-1/sinθ
斜率为-tanθ=tan(-θ)
所以,倾斜角为-θ
选A