If 4 sin2θ = 3(1 + cosθ), 0º < θ < 90°, then what is the value of (2 tanθ + 4sinθ - secθ)?
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Correct answer: B
Explanation
The correct answer is option 2 i.e. 3√15 - 4
4 sin2θ = 3(1 + cosθ)
4(1 - cos2θ) = 3(1 + cosθ)
4 - 4cos2θ = 3 + 3cosθ
4cos2θ + 3cosθ - 1 = 0
Let Cosθ = t
4t2 + 3t - 1 = 0
4t2 + 4t - t - 1 = 0
4t(t + 1) - 1(t + 1) = 0
(4t - 1)(t + 1) = 0
Cosθ = -1 or 1/4
We can't take Cosθ as -1 as its not in the range.
Cosθ = 1/4
Sin2θ = (1 - 1/16)
Sinθ = √(15)/4
Tanθ = √(15)
Secθ = 1/(1/4) = 4
2 tanθ + 4sinθ - secθ = 2(√15) + 4(√(15)/4) - 4
= 2√15 + √15 - 4
= 3√15 - 4