math 請教

2010-08-08 3:10 am
請問用甚麼方法去計算出答案的﹖

1/cos^2θ-1

1/cosθ-sinθtanθ

sinθ/√(1-cos^2θ)

1+tan^2θ

6sinθ/sin(90°-θ)

4cos(90°-θ)-cosθ/tan(90°-θ)

Thank you~
更新1:

本人為這小問題感到十分困惑,希望大家的協助

回答 (2)

2010-08-08 3:15 am
✔ 最佳答案
1/(cos^2θ-1)
= 1/sin^2 θ

1/cosθ-sinθtanθ
= 1/cosθ - sinθ * sinθ/cosθ
= (1-sin^2 θ)/cosθ
= cos^2 θ / cosθ
= cos θ

sinθ/√(1-cos^2θ)
= sinθ /√(sin^2 θ)
= sinθ/sinθ
= 1

1+tan^2θ<---For this:
(sin^2 θ + cos^2 θ = 1)
(tan^2 θ + 1 = 1/cos^2 θ)
(tan^2 θ + 1 = sec^2 θ)

6sinθ/sin(90°-θ)
= 6sinθ/cosθ
= 6tanθ

4cos(90°-θ)-cosθ/tan(90°-θ)
= 4 sinθ - cosθ / cotθ
= 4sinθ - cosθ * sinθ/cosθ
= 4sinθ - sinθ
= 3 sinθ
參考: Hope can help you^^”
2010-08-08 3:24 am
用計算機。。。。。。。。。。。


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