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若0<θ<π/2,化简(sinθ/1-cosθ)*根号tanθ-sinθ/tanθ+sinθ (大根号,后面都包括)
人气:496 ℃ 时间:2020-09-07 05:59:49
解答
[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
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