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
推荐
- 若cosα0,化简sinα+根号1-cos^2α
- 根号下sin²α(1+1/tanα)+cos²α(1+tanα)化简等于?
- 已知tanθ=根号下((1-a)/a),其中a大于0小于1 求:sin^2θ/(a+cosθ)+sin^2θ/(a-cosθ)的值.
- 化简:根号(1-cos^2α)-tanα根号(1-sin^2α),α(∏,2、3∏)
- 已知sinα-cosα=根号2,α∈(0,π),则tanα=
- 有一长方体木块,长30cm,宽20cm,高10cm.若把此木块锯成棱长是1cm的小正方体,并排成一长行,能排多长?
- 含有比喻修辞手法的成语 、 、
- 什么成语是即是褒义的又含有贬义那
猜你喜欢