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 $\left(\frac{l_1 l_2}{e_1 e_2}\right) \tan ^2 \theta$ is equal to