神马作文网求证[tanθ·(1-sinθ)]/1+cosθ=[cotθ·(1-cosθ)]/1+sinθ
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求证[tanθ·(1-sinθ)]/1+cosθ=[cotθ·(1-cosθ)]/1+sinθ
![求证[tanθ·(1-sinθ)]/1+cosθ=[cotθ·(1-cosθ)]/1+sinθ](/uploads/image/z/4473037-37-7.jpg?t=%E6%B1%82%E8%AF%81%5Btan%CE%B8%C2%B7%281-sin%CE%B8%29%5D%2F1%2Bcos%CE%B8%3D%5Bcot%CE%B8%C2%B7%281-cos%CE%B8%29%5D%2F1%2Bsin%CE%B8)
证明：因为：[tanθ·(1-sinθ)]/(1+cosθ) - [cotθ·(1-cosθ)]/(1+sinθ)
={[tanθ·(1-sinθ)]·(1+sinθ) - [cotθ·(1-cosθ)]·(1+cosθ)}/[(1+cosθ)(1+sinθ)]
={[tanθ·(1-sin²θ)] - [cotθ·(1-cos²θ)]}/[(1+cosθ)(1+sinθ)]
=(tanθ·cos²θ - cotθ·sin²θ)/[(1+cosθ)(1+sinθ)]
=(sinθ·cosθ - cosθ·sinθ)/[(1+cosθ)(1+sinθ)]
=0
所以：[tanθ·(1-sinθ)]/(1+cosθ) = [cotθ·(1-cosθ)]/(1+sinθ)
={[tanθ·(1-sinθ)]·(1+sinθ) - [cotθ·(1-cosθ)]·(1+cosθ)}/[(1+cosθ)(1+sinθ)]
={[tanθ·(1-sin²θ)] - [cotθ·(1-cos²θ)]}/[(1+cosθ)(1+sinθ)]
=(tanθ·cos²θ - cotθ·sin²θ)/[(1+cosθ)(1+sinθ)]
=(sinθ·cosθ - cosθ·sinθ)/[(1+cosθ)(1+sinθ)]
=0
所以：[tanθ·(1-sinθ)]/(1+cosθ) = [cotθ·(1-cosθ)]/(1+sinθ)
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