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

<p>$$s = \left\{ {0 \in \left( {0,{\pi \over 2}} \right):\sum\limits_{m = 1}^9 {\sec \left( {\theta + (m - 1){\pi \over 6}} \right)\sec \left( {\theta + {{m\pi } \over 6}} \right) = - {8 \over {\sqrt 3 }}} } \right\}$$.</p> <p>$$\sum\limits_{m = 1}^9 {{1 \over {\cos \left( {\theta + (m - 1){\pi \over 6}} \right)}}\cos \left( {\theta + m{\pi \over 6}} \right)} $$</p> <p>$${1 \over {\sin \left( {{\pi \over 6}} \right)}}\sum\limits_{m = 1}^9 {{{\sin \left[ {\left( {\theta + {{m\pi } \over 6}} \right) - \left( {\theta + (m - 1){\pi \over 6}} \right)} \right]} \over {\cos \left( {\theta + (m - 1){\pi \over 6}} \right)\cos \left( {\theta + m{\pi \over 6}} \right)}}} $$</p> <p>$$ = 2\sum\limits_{m = 1}^9 {\left[ {\tan \left( {\theta + {{m\pi } \over 6}} \right) - \tan \left( {\theta + (m - 1){\pi \over 6}} \right)} \right]} $$</p> <p>Now,</p> <p>$$\matrix{ {m = 1} & {2\left[ {\tan \left( {\theta + {\pi \over 6}} \right) - \tan (\theta )} \right]} \cr {m = 2} & {2\left[ {\tan \left( {\theta + {{2\pi } \over 6}} \right) - \tan \left( {\theta + {\pi \over 6}} \right)} \right]} \cr {\matrix{ . \cr . \cr . \cr } } & {} \cr {m = 9} & {2\left[ {\tan \left( {\theta + {{9\pi } \over 6}} \right) - \tan \left( {\theta + 8{\pi \over 6}} \right)} \right]} \cr } $$</p> <p>$\therefore$ $$ = 2\left[ {\tan \left( {\theta + {{3\pi } \over 2}} \right) - \tan \theta } \right] = {{ - 8} \over {\sqrt 3 }}$$</p> <p>$= - 2[\cot \theta + \tan \theta ] = {{ - 8} \over {\sqrt 3 }}$</p> <p>$$ = - {{2 \times 2} \over {2\sin \theta \cos \theta }} = {{ - 8} \over {\sqrt 3 }}$$</p> <p>$= {1 \over {\sin 2\theta }} = {2 \over {\sqrt 3 }}$</p> <p>$\Rightarrow \sin 2\theta = {{\sqrt 3 } \over 2}$</p> <p>$2\theta = {\pi \over 3}$</p> <p>$2\theta = {{2\pi } \over 3}$</p> <p>$\theta = {\pi \over 6}$</p> <p>$\theta = {\pi \over 3}$</p> <p>$\sum {{\theta _i} = {\pi \over 6} + {\pi \over 3} = {\pi \over 2}}$</p>