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

[2sin(θ-3π/2)cos(θ+π/2)-1]/1-2sin^2 θ
＝（－2scosθinθ－1）/cos2θ
再利用万能公式，这样比较直观
得到＝(-2tanθ-1-tan^2θ)/(1-tan^2θ)=(2tanθ+1+tan^2θ)/(tan^2θ-1)
[tan(9π+θ)+1]/tanθ-1
=(tanθ+1)/(tanθ-1 )=(2tanθ+1+tan^2θ)/(tan^2θ-1)
左边等于右边 得证
或者
[2sin(θ-3π/2)cos(θ+π/2)-1]/1-2sin^2 θ
＝-(sin2θ+1)/cos2θ
=-(cosθ+sinθ)^2/[(cosθ-sinθ)(cosθ+sinθ)]
=-(cosθ+sinθ)/(cosθ-sinθ)
=(tanθ+1)/(tanθ-1)
同样可以

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...