(a)find tanθ in term of k.
(b)Hence find 1/cosθ - tanθ.
更新1:
(b)Hence find 1/cos²θ - tan²θ.
Maths
2008-04-08 1:27 am
If cosθ=1/k,where k>1 and θlies in quadrant IV,
(a)find tanθ in term of k.
(b)Hence find 1/cosθ - tanθ.
(a)find tanθ in term of k.
(b)Hence find 1/cosθ - tanθ.
回答 (2)
2008-04-08 1:33 am
✔ 最佳答案
cosθ = 1/k, where k > 1,and θ lies in quadrant IV a. cos2θ = 1/k2
sec2θ = 1/cos2θ = k2
tan2θ = sec2θ - 1 = k2- 1
tanθ = -√(k2 - 1) (Since θ lies in quadrant IV)
b. 1/cosθ – tanθ
= k – [-√(k2 - 1)]
= k + √(k2 - 1)
2008-04-08 20:16:27 補充：
(b)Hence find 1/cos²θ - tan²θ.
1/cos²θ - tan²θ
= k² - (k² - 1)
= 1
參考: Myself~~~
2008-04-08 1:42 am
(a)
因為cosθ=1/k 姐係個三角條adj.side 係1 and 條斜邊係k 根據
pyth.them.條對邊就係開方1+k既2次方 so tanθ就等於開方1+k既2次方 over 1 姐係等於開方1+k既2次方
(b)
好易架姐 代番a題答案就得 答案係k減開方1+k既2次方
因為cosθ=1/k 姐係個三角條adj.side 係1 and 條斜邊係k 根據
pyth.them.條對邊就係開方1+k既2次方 so tanθ就等於開方1+k既2次方 over 1 姐係等於開方1+k既2次方
(b)
好易架姐 代番a題答案就得 答案係k減開方1+k既2次方
參考: mE...........................
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