唔該幫我做呢2條maths,thz"
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中二maths-三角比
2010-05-23 6:41 pm
回答 (1)
2010-05-23 8:16 pm
✔ 最佳答案
1.[4tanθ/cos(90º - θ)] - [1/sin(90º - θ)]
= [4tanθ/sinθ] - [1/cosθ]
= [4(sinθ/cosθ)/sinθ] - [1/cosθ]
= [4/cosθ] - [1/cosθ]
= 3/cosθ
(or 3secθ)
2.
L.H.S.
= tanθ + (1/tanθ)
= (sinθ/cosθ) + (cosθ/sinθ)
= (sin²θ/sinθcosθ) + (cos²θ/sinθcosθ)
= (sin²θ + cos²θ)/sinθcosθ
= 1/sinθcosθ
= 1/sinθsin(90º - θ)
= R.H.S.
參考: fooks
收錄日期: 2021-04-13 17:16:26
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100523000051KK00469