7. Given the parabola $y^{2}=4 x$, with its focus at $F$, a line passing through the focus $F$ and with an inclination angle of $\theta\left(0<\theta<\frac{\pi}{2}\right)$ intersects the parabola at points $A$ and $B$, $A O$ (where $O$ is the origin) intersects the directrix at point $B^{\prime}$, and $B O$ intersects the directrix at point $A^{\prime}$. Then the area of quadrilateral $A B B^{\prime} A^{\prime}$ is $\qquad$
7. $\frac{8}{\sin ^{3} \theta}$.

From the given conditions, we know that the slope of line $AB$ is $k=\tan \theta$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\left(y_{2}<0<y_{1}\right)$.
From $\left\{\begin{array}{l}y=k(x-1) \\ y^{2}=4 x\end{array}\right.$, we get
$y^{2}-\frac{4}{k} y-4=0$
$\Rightarrow y_{1}+y_{2}=\frac{4}{k}, y_{1} y_{2}=-4$.
Also, $l_{A O}: y=\frac{y_{1}}{x_{1}} x$, so $y=\frac{4}{y_{1}} x$.
Thus, the intersection point of $A O$ with the directrix is $B^{\prime}\left(-1,-\frac{4}{y_{1}}\right)$.
Since $y_{2}=-\frac{4}{y_{1}}$, the y-coordinates of points $B$ and $B^{\prime}$ are equal.
Therefore, $B B^{\prime} / / x$ axis.
Let the intersection point of $B O$ with the directrix be $A^{\prime}$.
Similarly, $A A^{\prime} / / x$ axis.
Thus, quadrilateral $A B B^{\prime} A^{\prime}$ is a right trapezoid.
Hence, $S_{\text {trapezoid } A B B^{\prime} A^{\prime}}=\frac{1}{2}\left(\left|A A^{\prime}\right|+\left|B B^{\prime}\right|\right)\left|A^{\prime} B^{\prime}\right|$
$$
\begin{array}{l}
=\frac{1}{2}|A B|\left|A^{\prime} B^{\prime}\right|=\frac{1}{2}\left(y_{2}-y_{1}\right)^{2} \sqrt{1+\frac{1}{k^{2}}} \\
=8\left(1+\frac{1}{k^{2}}\right)^{\frac{3}{2}}=8\left(1+\cot ^{2} \theta\right)^{\frac{3}{2}}=\frac{8}{\sin ^{3} \theta} .
\end{array}
$$