1. In a regular quadrilateral pyramid $P-ABCD$, all four lateral faces are equilateral triangles. Let the dihedral angle between a lateral face and the base be $\theta$, then $\tan \theta=$ $\qquad$
Let $P A=1, M$ be the midpoint of $A B$, and $O$ be the center of the square $A B C D$. Connect $P M, M O, O P$.

Since $P M \perp A B, O M \perp A B$, therefore, $\angle P M O$ is the dihedral angle formed by the side face $P A B$ and the base face $A B C D$.
Given $O M=\frac{1}{2}, P M=\frac{\sqrt{3}}{2}$, we have
$$
O P=\sqrt{P M^{2}-O M^{2}}=\frac{\sqrt{2}}{2} \text {. }
$$

Thus, $\tan \theta=\tan \angle P M O=\frac{O P}{O M}=\sqrt{2}$.