Question
If 2cosθ + 8sin2θ = 11, where
0o < θ < 90o, then find the value of cot θ.ÂSolution
Given, 2cosθ + 8sin2θ = 11 Or, 2cosθ + 8(1 – cos2θ) = 11 [Since, sin2θ = 1 – cos2θ] Or, 2cosθ + 8 – 8cos2θ = 11 Or, 8cos2θ – 2cosθ + 3 = 0 Or, 8cos2θ – 6cosθ + 4cosθ – 3 = 0 Or, 2cosθ(4cos – 3) + 1(4cosθ – 3) = 0 Or, (2cosθ + 1)(4cosθ – 3) = 0 Or, cosθ = (-1/2), (3/4) [(-1/2) is rejected as 0o < θ < 90o] We know that = cosθ = (Base)/(Hypotenuse) = (3/4) (Perpendicular)2 = (Hypotenuse)2 – (Base)2 (Perpendicular)2 = 42 – 32 = 16 – 9 = 7 Perpendicular = √7 So, cotθ = base/perpendicular = (3/√7)
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