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Question
If cosecθ + cotθ = 5, then evaluate secθ.
Solution
cosecθ + cotθ = 5
∴ `1/sinθ + cosθ/sinθ` = 5
∴ `(1 + cosθ)/sinθ = 5`
∴ 1 + cosθ = 5sinθ
Squaring both the sides, we get,
∴ (1 + cosθ)2 = 25sin2θ
∴ 1 + 2cosθ + cos2θ = 25sin2θ
∴ 1 + 2cosθ + cos2θ = 25(1 – cos2θ)
∴ 1 + 2cosθ + cos2θ = 25 – 25cos2θ
∴ 1 + 2cosθ + cos2θ + 25cos2θ – 25 = 0
∴ 26cos2θ + 2cosθ – 24 = 0
∴ 26cos2θ + 26cosθ − 24cosθ – 24 = 0
∴ 26cosθ (cosθ + 1) – 24(cosθ + 1) = 0
∴ (cosθ + 1)(26cosθ – 24) = 0
∴ cosθ + 1 = 0 or 26cosθ – 24 = 0
∴ cosθ = – 1 or cosθ = `24/26 = 12/13`
When cosθ = – 1, sinθ = 0
∴ cosecθ and cotθ are not defined.
∴ cosθ ≠ – 1
∴ cosθ = `12/13`
∴ secθ = `13/12`
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