若tan(π/4-θ)=3，则(cos2θ)/(1+sin2θ)=?

如果不要用万能公式怎么算

tan(π/4-θ)=(tanπ/4-tanθ)/(1+tanπ/4tanθ)
=(1-tanθ)/(1+tanθ)
(1-tanθ)/(1+tanθ)=3
1-tanθ=3(1+tanθ)
1-tanθ=3+3tanθ
4tanθ=-2
tanθ=-1/2
∵tanθ=-1/2
∴cosθ≠0
(cos2θ)/(1+sian2θ)
=(cos²θ-sin²θ)/(sin²θ+cos²θ+2sinθcosθ) 分子、分母同时乘1/cos²θ
=(1-sin²θ/cos²θ)/(sin²θ/cos²θ+1+2sinθ/cosθ)
=(1-tan²θ)/(tan²θ+1+2tanθ)
=[1-(-1/2)²]/[(-1/2)²+1+2×(-1/2)]
=(1-1/4)]/(1/4+1-1)
=(3/4)/(1/4)
=3

1+sin2θ-cos2θ/1+sin2θ+cos2θ
=(1-cos2θ)+sin2θ/(1+cos2θ)+sin2θ
=2(sinθ)^2+2sinθcosθ/2(cosθ)^2+2sinθcosθ
=sinθ/cosθ
=tanθ

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