求证 (sinθ+cosθ-1)(sinθ-cosθ+1)) /sin2θ=tanθ/2

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求证 (sinθ+cosθ-1)(sinθ-cosθ+1)) /sin2θ=tanθ/2
求证 (sinθ+cosθ-1)(sinθ-cosθ+1)) /sin2θ=tanθ/2

求证 (sinθ+cosθ-1)(sinθ-cosθ+1)) /sin2θ=tanθ/2
[(sinθ+cosθ-1)(sinθ-cosθ+1)]/sin2θ
=[(sinθ)^2-(cosθ-1)^2]/(2sinθcosθ)
=[(sinθ)^2-(cosθ)^2+2cosθ-1]/(2sinθcosθ)
=[2cosθ-2(cosθ)^2]/(2sinθcosθ)
=(1-cosθ)/sinθ
=[1-(1-2(sinθ/2)^2)]/(2sinθ/2cosθ/2)
=[2(sinθ/2)^2]/(2sinθ/2cosθ/2)
=sinθ/2/cosθ/2
=tanθ/2

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∵左边=[2sin(θ/2)cos(θ/2)-2sin^2(θ/2))(2sin(θ/2)cos(θ/2)+2sin^2(θ/2)]/sin2θ
=2sin(θ/2)[cos(θ/2)-sinθ/2]2sin(θ/2)[cos(θ/2)+sin(θ/2)]/sin2θ
=[4sin^2(θ/2)cosθ]/[2sinθcosθ]
=2sin^2(θ/2)/sinθ
=2sin^2(θ/2)/[2sin(θ/2)cos(θ/2)]
=sin(θ/2)/cos(θ/2)
=tanθ/2=右边
∴等式成立

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