If 4sin2θ = 3(1 + cosθ), 0° < θ < 90°, then what is the value of (2tanθ + 4sinθ - secθ)?

Answer: Option A

4sin²θ = 3(1 + cosθ)
4(1 - cos²θ) = 3 + 3cosθ
4 - 4cos²θ = 3 + 3cosθ
4cos²θ + 3cosθ - 1 = 0
4cos²θ + 4cosθ - cosθ - 1 = 0
4cosθ(cosθ + 1) - 1(cosθ + 1) = 0
(4cosθ -1)(cosθ + 1) = 0
4cosθ - 1 = 0
cosθ = $$\frac{1}{4}$$

Then, 2tanθ + 4tanθ - secθ
$$\begin{array}{rl}& =2\times \frac{\sqrt{15}}{1}+4\times \frac{\sqrt{15}}{4}-\frac{4}{1}\\ & =2\sqrt{15}+\sqrt{15}-4\\ & =3\sqrt{15}-4\end{array}$$