求证:(1-tanθ)/(1+tanθ)=(1-2sinθcosθ)/(cos2θ-sin2θ)
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求证:(1-tanθ)/(1+tanθ)=(1-2sinθcosθ)/(cos2θ-sin2θ)
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(1-tanθ)/(1+tanθ)=(cosθ-sinθ)/(cosθ+sinθ)=(cosθ-sinθ)^2/((cosθ+sinθ)*(cosθ-sinθ))=(cosθ^2-2sinθcosθ+sinθ^2)/(cos2θ-sin2θ)=(1-2sinθcosθ)/(cos2θ-sin2θ)
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:求证:(1-2sinθcosθ)/(cos2θ-sin2θ)=(cos2θ-sin2θ)/(1+2sinθcosθ).