数列 $\{a_n\}$ 满足 $a_1=\dfrac {\pi}{6}$，$a_{n+1}=\arctan (\sec a_n)$（$n \in \mathbb N^{\ast}$）．求正整数 $m$，使得$$\sin a_1 \cdot \sin a_2 \cdots \sin a_m=\dfrac {1}{100}.$$
答案 $3333$．
解析 根据题意，容易递推证明 $a_n\in\left(0,\dfrac{\pi}2\right)$，进而由题中递推公式可得\[\tan^2a_{n+1}=\tan^2a_n+1\implies \tan a_n=\sqrt{\dfrac{3n-2}3},\]因此\[\begin{split}\sin a_1 \cdot \sin a_2 \cdots \sin a_m&=\dfrac {\tan a_1}{\sec a_1}\cdot \dfrac {\tan a_2}{\sec a_2}\cdots \dfrac {\tan a_m}{\sec a_m}\\&= \dfrac {\tan a_1}{\tan a_2}\cdot \dfrac {\tan a_2}{\tan a_3}\cdots \dfrac {\tan a_m}{\tan a_{m+1}}\\&=\dfrac {\tan a_1}{\tan a_{m+1}}\\&=\sqrt {\dfrac {1}{3m+1}},\end{split}\]从而解得 $m=3333$．