Let 0 < θ < π/2. If the eccentricity of the hyperbola x²cos²(θ) - y²sin²(θ) = 1 is greater than 2, then the length of its latus rectum lies in the interval. (32,2] (1,32] (2,3] (3,∞)

The given equation of the hyperbola is x²cos²(θ) - y²sin²(θ) = 1.
The standard equation of a hyperbola is x²/a² - y²/b² = 1.
Comparing the given equation with the standard equation, we have a² = 1/cos²(θ) and b² = 1/sin²(θ).
Eccentricity e is given by e = √(1 + b²/a²) = √(1 + tan²(θ)) = sec(θ).
Given that e > 2, we have sec(θ) > 2, which means cos(θ) < 1/2.
Since 0 < θ < π/2, we have 0 < θ < π/3.
The length of the latus rectum is given by 2b²/a = 2(1/sin²(θ))/(1/cos²(θ)) = 2cot²(θ).
Since 0 < θ < π/3, we have cot²(θ) > 3/4.
Therefore, the length of the latus rectum is greater than 2(3/4) = 3/2 = 1.5.
Also, as θ approaches 0, the length of the latus rectum approaches infinity. Therefore, the length of the latus rectum lies in the interval (3/2, ∞).
The correct option is (3,∞).