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=(ta...

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