10. 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
$$
\sin a_{1} \cdot \sin a_{2} \cdots \sin a_{m}=\frac{1}{100} .
$$
Given $a_{n} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \Rightarrow \sec a_{n}>0 \Rightarrow a_{n+1} \in\left(0, \frac{\pi}{2}\right)$.
Then $a_{n+1}=\arctan \left(\sec a_{n}\right) \Rightarrow \tan a_{n+1}=\sec a_{n}=\sqrt{1+\tan ^{2} a_{n}}$
$\Rightarrow \tan ^{2} a_{n+1}-\tan ^{2} a_{n}=1$, and $\tan ^{2} a_{1}=\frac{1}{3}$,
Thus $\left\{\tan ^{2} a_{n}\right\}$ is an arithmetic sequence with the first term $\frac{1}{3}$ and common difference 1 $\Rightarrow \tan ^{2} a_{n}=\frac{3 n-2}{3}$.
Therefore, $\sin ^{2} a_{n}=\frac{3 n-2}{3 n+1} \Rightarrow \sin a_{n}=\sqrt{\frac{3 n-2}{3 n+1}}\left(n \in \mathbf{N}^{*}\right)$.
Hence, $\sin a_{1} \cdot \sin a_{2} \cdots \sin a_{m}=\sqrt{\frac{1}{4} \cdot \frac{4}{7} \cdots \frac{3 m-2}{3 m+1}}=\sqrt{\frac{1}{3 m+1}}=\frac{1}{100}$
$\Rightarrow 3 m+1=10000 \Rightarrow m=3333$