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

A. $$3\sqrt {15} - 4$$

B. $$15\sqrt 3 - 4$$

C. $$15\sqrt 3 + 3$$

D. $$4\sqrt {15} - 3$$

Answer: Option A

Solution(By Examveda Team)

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
$$\eqalign{ & = 2 \times \frac{{\sqrt {15} }}{1} + 4 \times \frac{{\sqrt {15} }}{4} - \frac{4}{1} \cr & = 2\sqrt {15} + \sqrt {15} - 4 \cr & = 3\sqrt {15} - 4 \cr} $$