证明(1-2sinθcosθ)/(cos^2θ-sin^2θ)=(cos^2θ-sin^2θ)/(1-2sinθcosθ
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证明(1-2sinθcosθ)/(cos^2θ-sin^2θ)=(cos^2θ-sin^2θ)/(1-2sinθcosθ)
证明:
左边=2sin(П+θ)cosθ-1/1-2sin^2θ
=(-2sinθcosθ-1)/cos2θ
=-(2sinθcosθ+sin^2 θ+cos^2 θ)/(cos^2 θ-sin^2 θ)
=-(sinθ+cosθ)^2/(cosθ-sinθ)(cosθ+sinθ)
=-(sinθ+cosθ)/(cosθ-sinθ)
=-[(sinθ/cosθ)+1]/[1-(sinθ/cosθ)]
=-(tanθ+1)/(1-tanθ)
=(tanθ+1)/(tanθ-1)
右边=tan(9П+θ)-1/tan(П+θ)+1
=(tanθ-1)/(tanθ+1)
左边=2sin(П+θ)cosθ-1/1-2sin^2θ
=(-2sinθcosθ-1)/cos2θ
=-(2sinθcosθ+sin^2 θ+cos^2 θ)/(cos^2 θ-sin^2 θ)
=-(sinθ+cosθ)^2/(cosθ-sinθ)(cosθ+sinθ)
=-(sinθ+cosθ)/(cosθ-sinθ)
=-[(sinθ/cosθ)+1]/[1-(sinθ/cosθ)]
=-(tanθ+1)/(1-tanθ)
=(tanθ+1)/(tanθ-1)
右边=tan(9П+θ)-1/tan(П+θ)+1
=(tanθ-1)/(tanθ+1)
化简:1+sinθ+cosθ+2sinθcosθ /1+sinθ+cosθ
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