Example 5 As shown in Figure $15-5$, in the prism $A B C-A_{1} B_{1} C_{1}$, $A B \perp$ side face $B B_{1} C_{1} C, E$ is a point on edge $C C_{1}$ different from $C$ and $C_{\text {: }}$. $E A \perp E B_{1}$. Given $A B=\sqrt{2}, B B_{1}=2 \cdot B C=1 \cdot \angle B C C_{1}=\frac{\pi}{3}$. Find the tangent value of the dihedral angle $A-E B_{\text {1 }}-A_{1}$.
Solution 1: Since $A B \perp$ plane $B B_{1} C_{1} C$, hence $A B \perp B E$.  
Also, $E B_{1} \perp E A$, and the projection of $E A$ in plane $B C C_{1} B_{1}$ is $E B$.  
By the inverse of the theorem of two perpendiculars, we know $E B_{1} \perp B E$.  
In parallelogram $B C C_{1} B_{1}$, let $E B=x$. Then $E B_{1}=\sqrt{4-x^{2}}$.  
As shown in Figure 15-6, draw $B D \perp C C_{1}$, intersecting $O C_{1}$ at $D$, then $B D=B C \cdot \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$.  

Solving, we get $x=1$ or $x=\sqrt{3}$.  
If $x=\sqrt{3}$, in $\triangle B C E$, $C E^{2}+1^{2}-2 C E \cdot \cos \frac{\pi}{3}=3$. Solving, we get $C E=2$.  
At this time, $E$ coincides with $C_{1}$, which is not feasible, so we discard it. Thus, $x=1, E B=1$.  
Draw $E G \perp B_{1} A_{1}$, then $E G \perp$ plane $B C C_{1} B_{1}$.  
Therefore, $G E \perp E B_{1}$ and $G E$ is in plane $A_{1} B_{1} E$.  
Also, by the given $A E \perp E B_{1}$, hence $\angle A E G$ is the plane angle of the dihedral angle $A-E B_{1}-A_{1}$.  
Since $E G / / B_{1} A_{1} / / B A, \angle A E G=\angle B A E$. Therefore, $\tan \angle A E G=\frac{B E}{A B}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$.  

Solution 2: As shown in Figure 15-7, with $B$ as the origin, $\overrightarrow{B B_{1}}, \overrightarrow{B A}$ as the $y$-axis, $z$-axis respectively, establish a spatial rectangular coordinate system, then $B(0,0,0), A(0,0, \sqrt{2}), B_{1}(0,2,0), C\left(\frac{\sqrt{3}}{2},-\frac{1}{2}, 0\right), C_{1}\left(\frac{\sqrt{3}}{2}, \frac{3}{2}, 0\right)$. Let $E\left(\frac{\sqrt{3}}{2}, a, 0\right)$, from $E A \perp E B_{1}$ we get $\overrightarrow{E A} \cdot \overrightarrow{E B_{1}}=0$.

That is, $0=\left(-\frac{\sqrt{3}}{2},-a, \sqrt{2}\right) \cdot\left(-\frac{\sqrt{3}}{2}, 2-a, 0\right)$
$$
=\frac{3}{4}+a(a-2)
$$

Solving, we get $a=\frac{1}{2}$ or $a=\frac{3}{2}$ (discard), hence $E\left(\frac{\sqrt{3}}{2}, \frac{1}{2}, 0\right)$.  
By the given, $\overrightarrow{E A} \perp \overrightarrow{E B_{1}}, \overrightarrow{B_{1} A_{1}} \perp \overrightarrow{E B_{1}}$,  
Therefore, the plane angle $\theta$ of the dihedral angle $A-E B_{1}-A_{1}$ is the angle between vectors $\overrightarrow{B_{1} A_{1}}$ and $\overrightarrow{E A}$.  
Since $\overrightarrow{B_{1} A_{1}}=\overrightarrow{B A}=(0,0, \sqrt{2}), \overrightarrow{E A}=\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}, \sqrt{2}\right)$  
Thus, $\cos \left\langle\overrightarrow{E A} \cdot \overrightarrow{B_{1} A_{1}}\right\rangle=\frac{\overrightarrow{E A} \cdot \overrightarrow{B_{1} \overrightarrow{A_{1}}}}{|\overrightarrow{E A}| \cdot\left|\overrightarrow{B_{1} A_{1}}\right|}=\frac{\sqrt{2}}{\sqrt{3}}$  
Hence, $\tan \left\langle\overrightarrow{E A}, \overrightarrow{B_{1} A_{1}}\right\rangle=\frac{\sqrt{2}}{2}$ is the required value.