If $cos θ = \frac{7}{3\sqrt{6}}$ and θ is an acute angle, then the value of $27 sin^2 θ - \frac{3}{2}$ is :

12 15 1 9

1

$cos θ = \frac{7}{3\sqrt{6}}$ cos²θ = \(\frac{49}{54}\) Now, 27sin²θ - \(\frac{3}{2}\) = 27 ( 1 - cos²θ ) - \(\frac{3}{2}\) = 27 ( 1 - \(\frac{49}{54}\)  ) - \(\frac{3}{2}\)  = 27 (  \(\frac{5}{54}\)  ) - \(\frac{3}{2}\)  = \(\frac{5}{2}\) - \(\frac{3}{2}\) = 1