证明(1-cos²θ/sinθ-cosθ)-(sinθ+cosθ/tan²θ-1)=sinθ+cosθ
问题描述:
证明(1-cos²θ/sinθ-cosθ)-(sinθ+cosθ/tan²θ-1)=sinθ+cosθ
答
证明:
左边=(1-cos²θ)/(sinθ-cosθ)-(sinθ+cosθ)/(tan²θ-1)
=sin²θ/(sinθ-cosθ)-(sinθ+cosθ)/(sin²θ/cos²θ-1)
=sin²θ/(sinθ-cosθ)-(sinθ+cosθ)cos²θ/(sin²θ-cos²θ)
=sin²θ/(sinθ-cosθ)-(sinθ+cosθ)cos²θ/[(sinθ+cosθ)(sinθ-cosθ)]
=sin²θ/(sinθ-cosθ)-cos²θ/(sinθ-cosθ)
=(sin²θ-cos²θ)/(sinθ-cosθ)
=(sinθ+cosθ)(sinθ-cosθ)/(sinθ-cosθ)
=sinθ+cosθ=右边
∴所证等式成立