10. As shown in Figure 2, in the tetrahedron $P-ABC$, $PA \perp$ the base $ABC$, $\angle ACB=90^{\circ}$, $AE \perp PB$ at $E$, $AF \perp PC$ at $F$. If $PA=AB=2$, $\angle BPC=\theta$, then when the area of $\triangle AEF$ is maximized, the value of $\tan \theta$ is ( ).
(A) 2
(B) $\frac{1}{2}$
(C) $\sqrt{2}$
(D) $\frac{\sqrt{2}}{2}$
10.D.

Since $P A \perp$ plane $A B C$, then $P A \perp B C$.
Also, $B C \perp A C$, so $B C \perp$ plane $P A C$.
Therefore, plane $P B C \perp$ plane $P A C$.
Since $A F \perp P C$, then $A F \perp$ plane $P B C$, hence $A F \perp E F$.
Also, $P A=A B=2, A E \perp P B$, so $A E=\sqrt{2}$.
In the right triangle $\triangle A E F$,
since $A F^{2}+E F^{2}=A E^{2}=(\sqrt{2})^{2}=2$,
we have $A F \cdot E F \leqslant \frac{A F^{2}+E F^{2}}{2}=1$.
Therefore, $S_{\triangle A E F}=\frac{1}{2} A F \cdot E F \leqslant \frac{1}{2} \times 1=\frac{1}{2}$.
The equality holds if and only if $A F=E F$, which means $S_{\triangle A E F}$ reaches its maximum value $\frac{1}{2}$.
At this time, $A E \cdot E F=E F^{2}=1 \Rightarrow E F=1$.
Since $A F \perp$ plane $P B C, A E \perp P B$, by the inverse of the three perpendiculars theorem, we get $F E \perp P B$.
In the right triangle $\triangle P E F$, given $P E=\sqrt{2}, E F=1$, we know $\tan \theta=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$.