求证：[2sin(θ-3π/2)cos(θ+π/2)-1]/1-2sin^2 θ=[tan(9π+θ)+1]/tanθ-

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求证：[2sin(θ-3π/2)cos(θ+π/2)-1]/1-2sin^2 θ=[tan(9π+θ)+1]/tanθ-1

左式=[2sin(θ+p/2)cos(θ+p/2)-1]/[1-2(sinθ)^2]=[sin(p+2θ)-1]/[1-2(sinθ)^2]
=-(sin2θ+1)/cos2θ=-(cosθ+sinθ)^2/[(cosθ-sinθ)(cosθ+sinθ)]
=-(cosθ+sinθ)/(cosθ-sinθ)=-(1+tanθ)/(1-tanθ)
右式=(tanθ+1)/(tanθ-1)=-(tanθ+1)/(1-tanθ)
所以：左式=右式
其中：p表示派

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