Mathematics Question from JEE Main 2026 · 28 January, Shift 1

For some $\theta\in\left(0,\frac{\pi}{2}\right)$, let the eccentricity and the length of the latus rectum of the hyperbola $x^2-y^2\sec^2\theta=8$ be $e_1$ and $l_1$, respectively, and let the eccentricity and the length of the latus rectum of the ellipse $x^2\sec^2\theta+y^2=6$ be $e_2$ and $l_2$, respectively. If $e_1^2=e_2^2\left(\sec^2\theta+1\right),$ then $\left(\frac{l_1l_2}{e_1e_2}\right)\tan^2\theta$ is equal to: