6. Given $\cos \beta+\sin \beta \cdot \cot \theta=\tan 54^{\circ}$, $\cos \beta-\sin \beta \cdot \cot \theta=\tan 18^{\circ}$.
Then the value of $\tan ^{2} \theta$ is $\qquad$
6.1

From the given conditions, we have
$\cos \beta=\frac{1}{2}\left(\tan 54^{\circ}+\tan 18^{\circ}\right)=\frac{1}{2 \cos 54^{\circ}}$,
$\sin \beta=\frac{\tan \theta}{2}\left(\tan 54^{\circ}-\tan 18^{\circ}\right)=\frac{\tan \theta}{2 \cos 18^{\circ}}$.
Since $\sin ^{2} \beta+\cos ^{2} \beta=1$, we have
$\frac{1}{\cos ^{2} 54^{\circ}}+\frac{\tan ^{2} \theta}{\cos ^{2} 18^{\circ}}=4$.
Thus, $\tan ^{2} \theta=\left(4-\frac{1}{\cos ^{2} 54^{\circ}}\right) \cos ^{2} 18^{\circ}$
$$
\begin{array}{l}
=4 \cos ^{2} 18^{\circ}-\frac{\sin ^{2} 72^{\circ}}{\sin ^{2} 36}=4 \cos ^{2} 18^{\circ}-4 \cos ^{2} 36^{\circ} \\
=4\left(\cos 18^{\circ}+\cos 36^{\circ}\right)\left(\cos 18^{\circ}-\cos 36^{\circ}\right)
\end{array}
$$
$$
\begin{array}{l}
=16 \cos 27^{\circ} \cdot \cos 9^{\circ} \cdot \sin 27^{\circ} \cdot \sin 9^{\circ} \\
=\frac{4 \sin 18^{\circ} \cdot \cos 18^{\circ} \cdot \sin 54^{\circ}}{\cos 18^{\circ}}=\frac{2 \sin 36^{\circ} \cdot \cos 36^{\circ}}{\cos 18^{\circ}} \\
=\frac{\sin 72^{\circ}}{\cos 18^{\circ}}=1 .
\end{array}
$$