Let $$0 < \theta < \frac{\pi}{2}$$. If the eccentricity of the hyperbola $$\frac{x^2}{\cos^2\theta} - \frac{y^2}{\sin^2\theta} = 1$$ is greater than 2, then the length of its latus rectum lies in the interval:

We are given the hyperbola $$\frac{x^{2}}{\cos^{2}\theta}-\frac{y^{2}}{\sin^{2}\theta}=1$$ with the condition $$0<\theta<\frac{\pi}{2}$$ and the extra information that its eccentricity is greater than 2. We must find the possible values of the length of its latus‐rectum.

For any hyperbola in the standard form $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$ we know two important formulas:

1. Eccentricity: $$e^{2}=1+\frac{b^{2}}{a^{2}}.$$
2. Length of the latus rectum: $$\text{latus rectum}= \frac{2b^{2}}{a}.$$

Comparing the given equation with the standard form, we can immediately read off

$$a^{2}=\cos^{2}\theta \quad\Longrightarrow\quad a=\cos\theta,$$

$$b^{2}=\sin^{2}\theta.$$

Now we apply the eccentricity formula. Substituting $$a^{2}=\cos^{2}\theta$$ and $$b^{2}=\sin^{2}\theta$$ into $$e^{2}=1+\dfrac{b^{2}}{a^{2}},$$ we obtain

$$e^{2}=1+\frac{\sin^{2}\theta}{\cos^{2}\theta}=1+\tan^{2}\theta.$$

By the Pythagorean identity, $$1+\tan^{2}\theta=\sec^{2}\theta,$$ so

$$e=\sec\theta.$$

The question states that the eccentricity is greater than 2, therefore

$$\sec\theta > 2\quad\Longrightarrow\quad \cos\theta <\frac12.$$

Because $$0<\theta<\frac{\pi}{2},$$ the cosine function decreases from 1 to 0 in this interval, so the inequality $$\cos\theta <\frac12$$ is equivalent to

$$\theta >\cos^{-1}\!\left(\frac12\right)=\frac{\pi}{3}.$$

Hence

$$\frac{\pi}{3}<\theta<\frac{\pi}{2}\quad\Longrightarrow\quad 0<\cos\theta<\frac12.$$

To determine the length of the latus rectum, we use the second formula stated above. Substituting $$b^{2}=\sin^{2}\theta$$ and $$a=\cos\theta,$$ we find

$$\text{latus rectum}= \frac{2\sin^{2}\theta}{\cos\theta}.$$

It is convenient to write everything in terms of $$x=\cos\theta,$$ because $$x$$ varies over the interval $$(0,\tfrac12).$$ Noting that $$\sin^{2}\theta=1-\cos^{2}\theta=1-x^{2},$$ we obtain

$$\text{latus rectum}=2\,\frac{1-x^{2}}{x}=2\!\left(\frac1x-x\right).$$

Define the function $$f(x)=2\!\left(\frac1x-x\right)$$ for $$x\in(0,\tfrac12).$$ We analyse how $$f(x)$$ behaves on this interval in order to find the possible range of the latus rectum.

The derivative is

$$f'(x)=2\!\left(-\frac1{x^{2}}-1\right)=-\frac{2}{x^{2}}-2,$$

which is strictly negative for all positive $$x.$$ Therefore $$f(x)$$ is strictly decreasing as $$x$$ increases. We now look at the two end‐points of the interval:

1. As $$x\to0^{+},$$ we have $$\frac1x\to\infty,$$ so $$f(x)\to\infty.$$

2. At the right‐hand end $$x=\frac12,$$ we get

$$f\!\left(\frac12\right)=2\!\left(\frac1{\frac12}-\frac12\right)=2\,(2-0.5)=2\,(1.5)=3.$$

Because $$f(x)$$ is decreasing, the values of the latus rectum fill exactly the open interval $$(3,\infty).$$

Hence, the correct answer is Option A.