Let $\mathrm{S}=\left\{\mathrm{\theta }\in \left(0,\frac{\mathrm{\pi }}{2}\right) : \sum _{\mathrm{m}=1}^9 \sec \left(\mathrm{\theta }+(\mathrm{m}-1)\frac{\mathrm{\pi }}{6}\right)\sec \left(\mathrm{\theta }+\frac{\mathrm{m\pi }}{6}\right)=-\frac{8}{\sqrt{3}}\right\}$. Then

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

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

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

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

              $\begin{array}{l}=2\sum _{\mathrm{m}=1}^9 \frac{\sin (\mathrm{\beta }-\mathrm{\alpha })}{\cos \mathrm{\alpha }\cdot \cos \mathrm{\beta }}=2\sum _{\mathrm{m}=1}^9 (\tan \mathrm{\beta }-\tan \mathrm{\alpha })\\ =2\sum _{\mathrm{m}=1}^9 \left(\tan \left(\mathrm{\theta }+\mathrm{m}\frac{\mathrm{\pi }}{6}\right)-\tan \left(\mathrm{\theta }+(\mathrm{m}-1)\frac{\mathrm{\pi }}{6}\right)\right)\\ =2\left(\tan \left(\mathrm{\theta }+\frac{9\mathrm{\pi }}{6}\right)-\tan \mathrm{\theta }\right)=2(-\cot \mathrm{\theta }-\tan \mathrm{\theta })=-\frac{8}{\sqrt{3}}\end{array}$

(Given)

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

$\Rightarrow \tan \mathrm{\theta }=\frac{1}{\sqrt{3}}$

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

$\sum _{\mathrm{\theta }\in \mathrm{S}} \mathrm{\theta }=\frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{3}=\frac{\mathrm{\pi }}{2}$