10. (This question is worth 20 points) The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{\pi}{6}, a_{n+1}=\arctan \left(\sec a_{n}\right)\left(n \in \mathbf{N}^{*}\right)$. Find the positive integer $m$, such that
$$
\sin a_{1} \cdot \sin a_{2} \cdots \cdot \sin a_{m}=\frac{1}{100} .
$$
Solution: From the given conditions, 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 $\sec a_{n}>0$, it follows that $a_{n+1} \in\left(0, \frac{\pi}{2}\right)$. From (1), we have $\tan ^{2} a_{n+1}=\sec ^{2} a_{n}=1+\tan ^{2} a_{n}$, hence
$$
\tan ^{2} a_{n}=n-1+\tan ^{2} a_{1}=n-1+\frac{1}{3}=\frac{3 n-2}{3} \text {, }
$$

which means $\tan a_{n}=\sqrt{\frac{3 n-2}{3}}$. $\qquad$
10 points
Therefore,
$$
\begin{aligned}
\sin a_{1} \cdot \sin a_{2} \cdots \cdots \sin a_{m} & =\frac{\tan a_{1}}{\sec a_{1}} \cdot \frac{\tan a_{2}}{\sec a_{2}} \cdots \cdot \frac{\tan a_{m}}{\sec a_{m}} \\
& \left.=\frac{\tan a_{1}}{\tan a_{2}} \cdot \frac{\tan a_{2}}{\tan a_{3}} \cdots \cdot \frac{\tan a_{m}}{\tan a_{m+1}} \text { (using (1) }\right) \\
& =\frac{\tan a_{1}}{\tan a_{m+1}}=\sqrt{\frac{1}{3 m+1}} .
\end{aligned}
$$

From $\sqrt{\frac{1}{3 m+1}}=\frac{1}{100}$, we get $m=3333$. $\qquad$