The given question:
$$\displaystyle \sum_{k=1}^{13} \dfrac{1}{\sin \left(\frac{\pi}{4} + \frac{(k-1)\pi}{6}\right) \sin \left(\frac{\pi}{4} + \frac{k\pi}{6}\right) }=?$$
By expanding the denominator using the $\sin (A+B)$ formula, I ended up with- $$\displaystyle \sum_{k=1}^{13} \dfrac{1}{\left(\frac{1}{\sqrt 2} \times \cos \frac{(k-1)\pi}{6} + \frac{1}{\sqrt 2} \times \sin \frac{(k-1)\pi}{6} \right)\left(\frac{1}{\sqrt 2} \times \cos \frac{k\pi}{6} + \frac{1}{\sqrt2}\times \sin \frac{k\pi}{6}\right)}$$ $$=\displaystyle \sum_{k=1}^{13} \dfrac{2}{\left(\cos \frac{(k-1)\pi}{6} + \sin \frac{(k-1)\pi}{6} \right)\left(\cos \frac{k\pi}{6} + \sin \frac{k\pi}{6}\right)}$$
We can expand the denominator and further simplify using the $\cos (A-B)$ and the $\sin (A+B)$ formulas, and after simplifying I have-
$$=\displaystyle \dfrac{4}{\sqrt 3}\sum_{k=1}^{13} \dfrac{1}{\sin(\frac{2k-1}{2}\frac{\pi}{6})}$$
The summation can be expanded as:
$$=\dfrac{4}{\sqrt 3} \left(\dfrac{1}{\sin \frac{\pi}{12}} + \dfrac{1}{\sin \frac{3\pi}{12}} + \dfrac{1}{\sin \frac{5\pi}{12}}+...+\dfrac{1}{\sin \frac{25\pi}{12}}\right)$$
How do I proceed from here?