Prove that cosec2θ - cot2θ ≡ tanθ
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Trigonometric Identities
2013-05-12 8:59 am
回答 (2)
2013-05-12 9:40 am
✔ 最佳答案
L.H.S.= cosec2θ - cot2θ
= (1 / sin2θ) - (cos2θ / sin2θ)
= [(sin²θ + cos²θ) / 2sinθcosθ] - [(cos²θ - sin²θ) / 2sinθcosθ]
= (sin²θ + cos²θ - cos²θ + sin²θ) / 2sinθcosθ
= 2sin²θ / 2sinθcosθ
= sinθ / cosθ
= tanθ
= R.H.S.
Hence, cosec2θ - cot2θ ≡ tanθ
2013-05-12 01:42:50 補充：
trigonometric identities used :
sin²θ + cos²θ = 1
cos2θ = cos²θ - sin²θ
sin2θ = 2sinθcosθ
參考: micatkie, micatkie
2013-05-12 10:40 pm
L.H.S.= cosec2θ - cot2θ
= (1 / sin2θ) - (cos2θ / sin2θ)
= [(sin²θ + cos²θ) / 2sinθcosθ] - [(cos²θ - sin²θ) / 2sinθcosθ]
= (sin²θ + cos²θ - cos²θ + sin²θ) / 2sinθcosθ
= 2sin²θ / 2sinθcosθ
= sinθ / cosθ
= tanθ
= R.H.S.
= (1 / sin2θ) - (cos2θ / sin2θ)
= [(sin²θ + cos²θ) / 2sinθcosθ] - [(cos²θ - sin²θ) / 2sinθcosθ]
= (sin²θ + cos²θ - cos²θ + sin²θ) / 2sinθcosθ
= 2sin²θ / 2sinθcosθ
= sinθ / cosθ
= tanθ
= R.H.S.
收錄日期: 2021-04-11 19:45:15
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