已知tanθ=2,则sin(π/2+θ)-cos(π-θ)÷sin(π/2-θ)-sin(π-θ)等于?

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已知tanθ=2,则sin(π/2+θ)-cos(π-θ)÷sin(π/2-θ)-sin(π-θ)等于?
已知tanθ=2,则sin(π/2+θ)-cos(π-θ)÷sin(π/2-θ)-sin(π-θ)等于?

已知tanθ=2,则sin(π/2+θ)-cos(π-θ)÷sin(π/2-θ)-sin(π-θ)等于?
[sin(π/2+θ)-cos(π-θ)]÷[sin(π/2-θ)-sin(π-θ)]=[cosθ+cosθ]/[cosθ-sinθ]=2/(1-tanθ)=-2

sin(π/2+θ)=cosθ
cos(π-θ)=-cosθ
sin(π/2-θ)=cosθ
sin(π-θ)=sinθ
sin(π/2+θ)-cos(π-θ)÷sin(π/2-θ)-sin(π-θ)
=(cosθ+cosθ)/(cosθ-sinθ)
=2cosθ/(cosθ-sinθ) 分子分母同时除以cosθ 得
=2/(1-tanθ) tanθ=2 代入
=2/(1-2)
=-2