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

$$\begin{gathered} \mathrm{Given} hyperbola \frac{{\mathrm{x}}^2}{{\cos }^2\mathrm{\theta }}-\frac{{\mathrm{y}}^2}{{\sin }^2\mathrm{\theta }}=1 \\ \mathrm{Given} \mathrm{e}>2\Rightarrow {\mathrm{e}}^2>4 \\ \Rightarrow \frac{{\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }}{{\cos }^2\mathrm{\theta }}>4 \\ \Rightarrow se{c}^2\theta >4 \\ \Rightarrow 1+{\tan }^2\mathrm{\theta }>4 \\ \Rightarrow {\tan }^2\mathrm{\theta }>3 \\ \Rightarrow \mathrm{tan\theta }>\sqrt{3} \\ \Rightarrow \mathrm{\theta }>\frac{\mathrm{\pi }}{3} \\ \Rightarrow \mathrm{\theta }\in \left(\frac{\mathrm{\pi }}{3}, \frac{\mathrm{\pi }}{2}\right) \\ \mathrm{Now} \mathrm{L}.\mathrm{L}.\mathrm{R}=\frac{2{\sin }^2\mathrm{\theta }}{\mathrm{cos\theta }} \\ =\frac{2(1-{\cos }^2\mathrm{\theta })}{\mathrm{cos\theta }} \\ \mathrm{L}.\mathrm{L}.\mathrm{R}=2(\mathrm{sec\theta }-\mathrm{cos\theta }) \\ \mathrm{If} \mathrm{\theta }=\frac{\mathrm{\pi }}{3}\Rightarrow \mathrm{L}.L.\mathrm{R}=2\left[2-\frac{1}{2}\right]=3 \\ \mathrm{If} \mathrm{\theta }=\frac{\mathrm{\pi }}{2}\Rightarrow \mathrm{L}.\mathrm{L}.\mathrm{R}=\infty \\ \therefore \mathrm{L}.\mathrm{L}.\mathrm{R}. \mathrm{lies} \mathrm{in} (3, \infty ) \end{gathered}$$