若0<θ<π/2,化简(sinθ/1-cosθ)*根号tanθ-sinθ/tanθ+sinθ (大根号,后面都包括)

S欣蓝L 1年前 已收到1个回答 举报

aflush 幼苗

共回答了16个问题采纳率:75% 举报

[sinθ/(1-cosθ)]•√[(tanθ-sinθ)/(tanθ+sinθ)]
= [sinθ/(1-cosθ)]•√[(tanθ-tanθ•cosθ)/(tanθ+tanθ•cosθ)]
= [sinθ/(1-cosθ)]•√[(1-cosθ)/(1+cosθ)]
= [sinθ/(1-cosθ)]•√[(1-cosθ)²/(1+cosθ)(1-cosθ)]
=[sinθ/(1-cosθ)]•√[(1-cosθ)²/(1-cos²θ)]
=[sinθ/(1-cosθ)]•[(1-cosθ)/sinθ]
=1

1年前

3
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