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_____

The given hyperbola is $$x^{2}-y^{2}\sec^{2}\theta = 8$$.

Rewrite it in standard form:
$$x^{2}-\sec^{2}\theta\,y^{2}=8$$
$$\Longrightarrow\; \frac{x^{2}}{8}-\frac{y^{2}}{8/\sec^{2}\theta}=1.$$

Hence for the hyperbola we have
$$a^{2}=8,\qquad b^{2}=8\cos^{2}\theta.$$

Formula for eccentricity of $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$ is
$$e_{1}^{2}=1+\frac{b^{2}}{a^{2}}.$$

Using the above values:
$$e_{1}^{2}=1+\frac{8\cos^{2}\theta}{8}=1+\cos^{2}\theta\;.$$

Formula for length of the latus-rectum of a hyperbola is
$$l_{1}= \frac{2b^{2}}{a}.$$

Here $$a=\sqrt{8}=2\sqrt{2}$$, so
$$l_{1}= \frac{2\,(8\cos^{2}\theta)}{2\sqrt{2}} =\frac{16\cos^{2}\theta}{2\sqrt{2}} =4\sqrt{2}\,\cos^{2}\theta.$$

The given ellipse is $$x^{2}\sec^{2}\theta + y^{2}=6.$$

Rewrite it in standard form:
$$\frac{x^{2}\sec^{2}\theta}{6}+\frac{y^{2}}{6}=1 \;\Longrightarrow\; \frac{x^{2}}{6\cos^{2}\theta}+\frac{y^{2}}{6}=1.$$

Comparing with $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ (where $$b\gt a$$ because $$6\gt 6\cos^{2}\theta$$):
$$a^{2}=6\cos^{2}\theta,\qquad b^{2}=6.$$

Formula for eccentricity of an ellipse ($$b\gt a$$) is
$$e_{2}^{2}=1-\frac{a^{2}}{b^{2}}.$$

Therefore
$$e_{2}^{2}=1-\frac{6\cos^{2}\theta}{6}=1-\cos^{2}\theta=\sin^{2}\theta.$$

For an ellipse the length of the latus-rectum is
$$l_{2}=\frac{2a^{2}}{b}\;.$$

Here $$a^{2}=6\cos^{2}\theta$$ and $$b=\sqrt{6}$$, so
$$l_{2}= \frac{2\,(6\cos^{2}\theta)}{\sqrt{6}} =\frac{12\cos^{2}\theta}{\sqrt{6}} =2\sqrt{6}\,\cos^{2}\theta.$$

The condition given in the problem is
$$e_{1}^{2}=e_{2}^{2}\left(\sec^{2}\theta+1\right).$$

Substitute $$e_{1}^{2}=1+\cos^{2}\theta$$ and $$e_{2}^{2}=\sin^{2}\theta$$:
$$1+\cos^{2}\theta=\sin^{2}\theta\,(\sec^{2}\theta+1).$$

But $$\sec^{2}\theta+1=\frac{1+\cos^{2}\theta}{\cos^{2}\theta}$$, so
$$1+\cos^{2}\theta=\sin^{2}\theta\,\frac{1+\cos^{2}\theta}{\cos^{2}\theta}.$$

Divide both sides by $$1+\cos^{2}\theta\;( \gt 0)$$:
$$1=\frac{\sin^{2}\theta}{\cos^{2}\theta}=\tan^{2}\theta.$$

Thus $$\tan^{2}\theta=1\;\Longrightarrow\;\theta=\frac{\pi}{4}$$ (since $$0\lt\theta\lt\frac{\pi}{2}$$).

Compute the required quantities at $$\theta=\frac{\pi}{4}$$:
$$\cos\theta=\frac{1}{\sqrt{2}},\quad\sin\theta=\frac{1}{\sqrt{2}},\quad\tan^{2}\theta=1.$$

1. Eccentricities
$$e_{1}= \sqrt{1+\cos^{2}\theta}= \sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}} =\frac{\sqrt{3}}{\sqrt{2}},$$
$$e_{2}= \sin\theta=\frac{1}{\sqrt{2}}.$$

2. Lengths of latus-rectum
$$l_{1}=4\sqrt{2}\cos^{2}\theta =4\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)^{2} =4\sqrt{2}\left(\frac{1}{2}\right)=2\sqrt{2},$$
$$l_{2}=2\sqrt{6}\cos^{2}\theta =2\sqrt{6}\left(\frac{1}{2}\right)=\sqrt{6}.$$

Now evaluate the requested expression:
$$\frac{l_{1}l_{2}}{e_{1}e_{2}}\tan^{2}\theta =\left(\frac{2\sqrt{2}\;\cdot\;\sqrt{6}} {\left(\frac{\sqrt{3}}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)}\right)\,(1).$$

Simplify step by step:
Numerator $$l_{1}l_{2}=2\sqrt{2}\cdot\sqrt{6}=2\sqrt{12}=4\sqrt{3},$$
Denominator $$e_{1}e_{2}= \frac{\sqrt{3}}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}} =\frac{\sqrt{3}}{2}.$$
Therefore
$$\frac{l_{1}l_{2}}{e_{1}e_{2}} =\frac{4\sqrt{3}}{\sqrt{3}/2}=4\sqrt{3}\cdot\frac{2}{\sqrt{3}}=8.$$

Multiplying by $$\tan^{2}\theta=1$$ leaves the value unchanged.

Hence $$\left(\frac{l_{1}l_{2}}{e_{1}e_{2}}\right)\tan^{2}\theta = 8.$$

Final Answer: 8