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

$<blockquote><p>Let$\theta + \left(m - 1\right)\dfrac{π}{6} = x$and$\theta + m\dfrac{π}{6} = y$</p><p>So,$y - x = \dfrac{π}{6}$</p><p>Now,$\mathop{∑}_{m = 1}^{9}\sec \left(\theta + \left(m - 1\right)\dfrac{π}{6}\rrright)\sec \left(\theta + \dfrac{m\pi}{6}\right)</p><p>= \mathop{∑}_{m = 1}^{9}\sec x \sec y = \mathop{∑}_{m = 1}^{9}\dfrac{1}{\cos x \cos y}</p><p>= 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{π}{6}\right) -\tan \left(\theta + ($m - 1 )$\dfrac{π}{6}\right)\right)

= 2\left(\tan \left(\theta + \dfrac{π}{6}\right) -\tan \left(\theta + \dfrac{0 \cdot \pi}{6}\right)\right) +2 \left(\tan \left(\theta +2\dfrac{π}{6}\right) -\tan \left(\theta +\dfrac{π}{6}\right)\right)$</p><p>$+ 2\left(\tan \left(\theta +3\dfrac{π}{6}\right) -\tan \left(\theta +2\dfrac{π}{6}\right)\right) + . . . +2 \left(\tan \left(\theta +9\dfrac{π}{6}\right) -\tan \left(\theta +8\dfrac{π}{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}{3} (Given)</p><p>∴ tan$\theta +$cot$\theta = \dfrac{4}{3}</p><p>$⇒ tan\theta = \dfrac{1}{3}or \sqrt{3} (as \theta \in \left(0, \dfrac{π}{2}\right))$</p><p>So,$S = \left(\{\dfrac{π}{6}, \dfrac{π}{3}\right\}

Hence, $\mathop{∑}_{\theta \iinS}\theta = \dfrac{π}{6} + \dfrac{π}{3} = \dfrac{π}{2}$

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