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(A) Given the hyperbola equation: $\frac{x^2}{\cos^2 	heta} - \frac{y^2}{\sin^2 	heta} = 1$.Here,$a^2 = \cos^2 	heta$ and $b^2 = \sin^2 	heta$.The

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$(3, \infty)$

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(A) Given the hyperbola equation: $\frac{x^2}{\cos^2 	heta} - \frac{y^2}{\sin^2 	heta} = 1$.Here,$a^2 = \cos^2 	heta$ and $b^2 = \sin^2 	heta$.The eccentricity $e$ is given by $e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{\sin^2 	heta}{\cos^2 	heta} = 1 + 	an^2 	heta = \sec^2 	heta$.Given $e > 2$,so $e^2 > 4$,which implies $\sec^2 	heta > 4$,or $	an^2 	heta > 3$.Since $0  \sqrt{3}$,which means $	heta \in (\frac{\pi}{3}, \frac{\pi}{2})$.The length of the latus rectum $L$ is given by $L = \frac{2b^2}{a} = \frac{2 \sin^2 	heta}{\cos 	heta} = 2 	an 	heta \sin 	heta$.As $	heta$ increases from $\frac{\pi}{3}$ to $\frac{\pi}{2}$,$	an 	heta$ increases from $\sqrt{3}$ to $\infty$ and $\sin 	heta$ increases from $\frac{\sqrt{3}}{2}$ to $1$.Thus,$L = 2 	an 	heta \sin 	heta > 2(\sqrt{3})(\frac{\sqrt{3}}{2}) = 3$.As $	heta 	o \frac{\pi}{2}$,$L 	o \infty$.Therefore,the length of the latus rectum lies in the interval $(3, \infty)$.

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

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