23. The line $l$ passing through point $E\left(-\frac{p}{2}, 0\right)$ intersects the parabola $C: y^{2}=2 p x(p>0)$ at points $A$ and $B$, and the inclination angle of line $l$ is $\alpha$. Then the range of $\alpha$ is $\qquad$; $F$ is the focus of the parabola, and the area of $\triangle A B F$ is $\qquad$ (expressed in terms of $p$ and $\alpha$).
23. $\left(0, \frac{\pi}{4}\right) \cup\left(\frac{3 \pi}{4}, \pi\right), \frac{p^{2} \sqrt{\cos 2 \alpha}}{\sin \alpha}$.

Let the equation of line $l$ be $y=k\left(x+\frac{p}{2}\right)$.
Solving the system of equations $\left\{\begin{array}{l}y=k\left(x+\frac{p}{2}\right) \\ y^{2}=2 p x\end{array}\right.$, we get
$$
k^{2} x^{2}+\left(k^{2} p-2 p\right) x+\frac{k^{2} p^{2}}{4}=0 \text {. }
$$

From $\Delta>0$, we solve to get $-1<k<1$ and $k \neq 0$, which gives the range of the inclination angle.
$$
\begin{array}{l}
S_{\triangle A B Y}=\left|S_{\triangle A B F}-S_{\triangle A E F}\right|=\frac{1}{2} p\left|y_{B}-y_{A}\right| \\
=\frac{1}{2} p|k|\left|x_{B}-x_{A}\right|=p^{2} \sqrt{\frac{1-k^{2}}{k^{2}}}=\frac{p^{2} \sqrt{\cos 2 \alpha}}{\sin \alpha} .
\end{array}
$$