Let S = \ θ ∈ (0, (π)/(2)): ∑ _ m = 1 ^ 9 (θ + (m - 1)(π)/(6)) (θ +...

Let $\theta + \left(m - 1\right)\dfrac{\pi}{6} = x$ and $\theta + m\dfrac{\pi}{6} = y$

So, $y - x = \dfrac{\pi}{6}$

Now, $\mathop{∑}_{m = 1}^{9}\sec \left(\theta + \left(m - 1\right)\dfrac{\pi}{6}\right)\sec \left(\theta + \dfrac{m\pi}{6}\right)$

$= \mathop{∑}_{m = 1}^{9}\sec x \sec y = \mathop{∑}_{m = 1}^{9}\dfrac{1}{\cos x \cos y}$

$= 2\mathop{∑}_{m = 1}^{9}\dfrac{\sin \left(y - x\right)}{\cos x \cos y} = 2\mathop{∑}_{m = 1}^{9}\left(\tan y - \tan x\right)$

$= 2\mathop{∑}_{m = 1}^{9}\left(\tan \left(\theta + m\dfrac{\pi}{6}\right) - \tan \left(\theta + ( m - 1 ) \dfrac{\pi}{6}\right)\right)$

$= 2\left(\tan \left(\theta + \dfrac{\pi}{6}\right) - \tan \left(\theta + \dfrac{0 \cdot \pi}{6}\right)\right) + 2\left(\tan \left(\theta + 2\dfrac{\pi}{6}\right) - \tan \left(\theta + \dfrac{\pi}{6}\right)\right)$

$+ 2\left(\tan \left(\theta + 3\dfrac{\pi}{6}\right) - \tan \left(\theta + 2\dfrac{\pi}{6}\right)\right) + . . . + 2\left(\tan \left(\theta + 9\dfrac{\pi}{6}\right) - \tan \left(\theta + 8\dfrac{\pi}{6}\right)\right)$

$= 2\left(\tan \left(\theta + \dfrac{9\pi}{6}\right) - tan\theta\right) = 2\left(- cot\theta - tan\theta\right)$

i.e. $2\left(- cot\theta - tan\theta\right) = - \dfrac{8}{\sqrt{3}}$ (Given)

$∴ tan\theta + cot\theta = \dfrac{4}{\sqrt{3}}$

$⇒ tan\theta = \dfrac{1}{\sqrt{3}}$ or $\sqrt{3}$ (as $\theta \in \left(0, \dfrac{\pi}{2}\right)$)

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

Hence, $\mathop{∑}_{\theta \in S}\theta = \dfrac{\pi}{6} + \dfrac{\pi}{3} = \dfrac{\pi}{2}$