Five. (20 points) In the Cartesian coordinate system, $O$ is the origin, $A$ and $B$ are points in the first quadrant, and $A$ lies on the line $y = (\tan \theta) x$, where $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right), |OA| = \frac{1}{\sqrt{2} - \cos \theta}, B$ is a point on the hyperbola $x^{2} - y^{2} = 1$ that minimizes the area of $\triangle OAB$. Find the value of $\theta$ for which the area of $\triangle OAB$ is maximized, and determine the maximum area.

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As shown in Figure 6, draw the tangent line $B C$ of the hyperbola through point $B$, and $B C \parallel A O$.
Then $l_{B C}$:
$y=(\tan \theta) x+m$.
Substitute into the hyperbola equation $x^{2}-y^{2}=1$, and by $\Delta=0$ we get
$m^{2}=\tan ^{2} \theta-1$.
Also, $l_{A O}: y=(\tan \theta) x$,
$S_{\triangle O A B}=\frac{1}{2}|O A| \cdot h$ (where $h$ is the height on $O A$).
From equations (1), (2), and the distance formula between parallel lines, we get
$h=\frac{\sqrt{\tan ^{2} \theta-1}}{\sqrt{\tan ^{2} \theta+1}}$.
Thus, $S_{\triangle O A B}=\frac{1}{2} \cdot \frac{\sqrt{\tan ^{2} \theta-1}}{\sqrt{\tan ^{2} \theta+1}} \cdot \frac{1}{\sqrt{2}-\cos \theta}$
$$
=\frac{1}{2} \frac{\sqrt{\sec ^{2} \theta-2}}{\sec \theta} \cdot \frac{1}{\sqrt{2}-\cos \theta} \text {. }
$$

Let $\cos \theta=t$. Then,
$$
z=\frac{\sqrt{\sec ^{2} \theta-2}}{\sec \theta} \cdot \frac{1}{\sqrt{2}-\cos \theta} \text {. }
$$

Simplifying and rearranging, we get
$$
\left(z^{2}+2\right) t^{2}-2 \sqrt{2} z^{2} t+2 z^{2}-1=0 \text {. }
$$

Then $\Delta^{\prime}=8 z^{4}-4\left(z^{2}+2\right)\left(2 z^{2}-1\right) \geqslant 0$.
Solving, we get $z \leqslant \frac{\sqrt{6}}{3}$.
When $\cos \theta=\frac{\sqrt{2}}{4}$, i.e., $\theta=\arccos \frac{\sqrt{2}}{4} \in$ $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, $S_{\triangle O A B}$ has a maximum value of $\frac{\sqrt{6}}{6}$.