10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{\pi}{6}, a_{n+1}=\arctan \left(\sec a_{n}\right)\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Find the positive integer $m$ such that
$$
\sin a_{1} \cdot \sin a_{2} \cdots \cdot \sin a_{m}=\frac{1}{100} .
$$
10. From the problem, we know that for any positive integer $n$,
$$
a_{n+1} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right),
$$

and $\tan a_{n+1}=\sec a_{n}$.
Since $\sec a_{n}>0$, then $a_{n+1} \in\left(0, \frac{\pi}{2}\right)$.
From equation (1), we get $\tan ^{2} a_{n+1}=\sec ^{2} a_{n}=1+\tan ^{2} a_{n}$.
Thus, $\tan ^{2} a_{n}=n-1+\tan ^{2} a_{1}$
$$
\begin{array}{l}
=n-1+\frac{1}{3}=\frac{3 n-2}{3} \\
\Rightarrow \tan a_{n}=\sqrt{\frac{3 n-2}{3}} .
\end{array}
$$

Therefore, $\sin a_{1} \cdot \sin a_{2} \cdots \cdot \sin a_{m}$
$=\frac{\tan a_{1}}{\sec a_{1}} \cdot \frac{\tan a_{2}}{\sec a_{2}} \cdots \cdots \cdot \frac{\tan a_{m}}{\sec a_{m}}$
$=\frac{\tan a_{1}}{\tan a_{2}} \cdot \frac{\tan a_{2}}{\tan a_{3}} \cdots \cdot \frac{\tan a_{m}}{\tan a_{m+1}}$ (using equation (1))
$=\frac{\tan a_{1}}{\tan a_{m+1}}=\sqrt{\frac{1}{3 m+1}}$.
From $\sqrt{\frac{1}{3 m+1}}=\frac{1}{100} \Rightarrow m=3333$.