If $\frac{3\sqrt{3}secθ+4 tan θ}{3 tan θ + \sqrt{3} sec θ}=2, 0° < θ < 90°$, then the value of cos θ will be :

$\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{1}{4}$ $\frac{1}{2}$

$\frac{1}{2}$

$\frac{3\sqrt{3}secθ+4 tan θ}{3 tan θ + \sqrt{3} sec θ}=2, 0° < θ < 90°$ 3√3 secθ + 4 tanθ = 6 tanθ + 2√3 secθ √3 secθ  = 2 tanθ cosec θ = \(\frac{2}{√3}\) { we know, cosec 60º = \(\frac{2}{√3}\) } Now, cosθ = cos 60º = \(\frac{1}{2}\)