Let $\alpha=\theta+(m-1) \frac{\pi}{6}$

$\beta=\theta+m \frac{\pi}{6}$

So, $\beta-\alpha=\frac{\pi}{6}$

Here,$\sum_{m=1}^{9} \sec \alpha \cdot \sec \beta=\sum_{m=1}^{9} \frac{1}{\cos \alpha \cdot \cos \beta}$

$= 2 \sum_{m=1}^{9} \frac{\sin (\beta-\alpha)}{\cos \alpha \cdot \cos \beta}=2 \sum_{m=1}^{9}(\tan \beta-\tan \alpha)$

$= 2 \sum_{m=1}^{9}\left(\tan \left(\theta+m \frac{\pi}{6}\right)-\tan \left(\theta+(m-1) \frac{\pi}{6}\right)\right)$

$=2\left(\tan \left(\theta+\frac{9 \pi}{6}\right)-\tan \theta\right)=2(-\cot \theta-\tan \theta)=-\frac{8}{\sqrt{3}}$

(Given)

$\therefore \quad \tan \theta+\cot \theta=\frac{4}{\sqrt{3}}$

$\tan \theta=\frac{1}{\sqrt{3}}$ or $\sqrt{3}$

So, $S=\left\{\frac{\pi}{6}, \frac{\pi}{3}\right\}$

$\sum_{\theta \in S} \theta=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}$