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secondary 4 | A Maths
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Date Posted: 4 years ago
Views: 252
(1 - sin2θ)/(1 + cos2θ)
= (1 - 2sinθcosθ)/(1 + 2cos²θ - 1)
= (1 - 2sinθcosθ)/(2cos²θ)
= ½ (1/cos²θ - 2sinθ/cosθ)
= ½( sec²θ - 2tanθ)
= ½(1 - 2tanθ + tan²θ)
= ½(1 - tan²θ) (proved)
= (1 - 2sinθcosθ)/(1 + 2cos²θ - 1)
= (1 - 2sinθcosθ)/(2cos²θ)
= ½ (1/cos²θ - 2sinθ/cosθ)
= ½( sec²θ - 2tanθ)
= ½(1 - 2tanθ + tan²θ)
= ½(1 - tan²θ) (proved)
Wow, how did I not think of that
When you see that the right hand side expression has no fraction, first thing you should try to do is to get rid of the denominator.
There are two terms in the denominator so try to reduce it to one. Then numerator is divided by it , thereby removing the denominator.
At the same time, there's 2θ in the terms so try to change to in terms of θ
There are two terms in the denominator so try to reduce it to one. Then numerator is divided by it , thereby removing the denominator.
At the same time, there's 2θ in the terms so try to change to in terms of θ
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done
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Hope this helps! I worked from tbe right hand side via expansion instead of working from the left via factorisation
Date Posted:
4 years ago
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