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secondary 3 | A Maths
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Part b
Date Posted: 4 years ago
Views: 392
(sin³θ + cos³θ)/(2sin²θ - 1)
= (sinθ + cosθ)(sin²θ -sinθcosθ + cos²θ)/(2sin²θ - (sin²θ + cos²θ))
= (sinθ + cosθ)(1 - sinθcosθ)/(sin²θ - cos²θ)
= (sinθ + cosθ)(1 - sinθcosθ)/[(sinθ + cosθ)(sinθ - cosθ)]
= (1 - sinθcosθ)/(sinθ - cosθ)
= (1/cosθ - sinθ)/(sinθ/cosθ - 1)
(Divide both numerator and denominator by cosθ)
= (secθ - sinθ)/(tanθ - 1)
= (sinθ + cosθ)(sin²θ -sinθcosθ + cos²θ)/(2sin²θ - (sin²θ + cos²θ))
= (sinθ + cosθ)(1 - sinθcosθ)/(sin²θ - cos²θ)
= (sinθ + cosθ)(1 - sinθcosθ)/[(sinθ + cosθ)(sinθ - cosθ)]
= (1 - sinθcosθ)/(sinθ - cosθ)
= (1/cosθ - sinθ)/(sinθ/cosθ - 1)
(Divide both numerator and denominator by cosθ)
= (secθ - sinθ)/(tanθ - 1)
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Date Posted:
4 years ago
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