Independent of $k$

$\begin{aligned} S & = \sum_{k=1}^{13} \frac 1{\sin\left(\frac \pi 4 + \frac {(k-1)\pi}6 \right)\sin\left(\frac \pi 4 + \frac {k\pi}6 \right)} \\ & = \sum_{k=1}^{13} \frac 1{\sin \frac {(2k+1)\pi}{12} \sin \frac {(2k+3)\pi}{12}} \\ & = \sum_{\color{#D61F06}k=2}^{\color{#D61F06}14} \frac 1{\sin \frac {({\color{#D61F06}2k-1})\pi}{12} \sin \frac {({\color{#D61F06}2k+1})\pi}{12}} \\ & = \sum_{k=2}^{14} \frac 2{\cos \frac \pi 6 - \cos \frac {k\pi}3} \end{aligned}$

Let us now consider the sum of six consecutive terms or $k = 2$ to $k =7$ . Since $\cos \frac {7\pi}3 = \cos \frac \pi 3$ , it is similar to considering $k = 1$ to $k =6$ .

$\begin{aligned} S_6 & = \sum_{k=1}^6 \frac 2{\cos \frac \pi 6 - \cos \frac {k\pi}3} \\ & = \frac 2{\frac {\sqrt 3}2 - \frac 12} + \frac 2{\frac {\sqrt 3}2 + \frac 12} + \frac 2{\frac {\sqrt 3}2 +1} + \frac 2{\frac {\sqrt 3}2 + \frac 12} + \frac 2{\frac {\sqrt 3}2 - \frac 12} + \frac 2{\frac {\sqrt 3}2 -1} \\ & = \frac 4{\frac {\sqrt 3}2 - \frac 12} + \frac 4{\frac {\sqrt 3}2 + \frac 12} + \frac 2{\frac {\sqrt 3}2 +1} + \frac 2{\frac {\sqrt 3}2 -1} \\ & = \frac {4 \left( \frac {\sqrt 3}2 + \frac 12 + \frac {\sqrt 3}2 - \frac 12\right)}{\left(\frac {\sqrt 3}2 - \frac 12\right)\left(\frac {\sqrt 3}2 + \frac 12\right)} + \frac {2 \left( \frac {\sqrt 3}2 - 1 + \frac {\sqrt 3}2 + 1 \right)}{\left(\frac {\sqrt 3}2 + 1 \right)\left(\frac {\sqrt 3}2 - 1 \right)} \\ & = \frac {4\sqrt 3}{\frac 12} - \frac {2\sqrt 3}{\frac 14} = 8 \sqrt 3 - 8 \sqrt 3 = 0 \end{aligned}$

Since the summand is periodical, the sum of every six consecutive terms is equal to 0. As we need to sum 13 terms the sum of the first 12 terms is 0, therefore the sum $S$ is equal to the last term:

$\begin{aligned} S & = \frac 2{\cos \frac \pi 6 - \cos \frac {14\pi}3} = \frac 2{\cos \frac \pi 6 - \cos \frac {2\pi}3} = \frac 2{\frac {\sqrt 3}2 + \frac 12} = \boxed{2(\sqrt 3-1)} \end{aligned}$

$\large Multiply\quad and\quad divide\quad by\quad \large{ \sin { \frac { \pi }{ 6 } } \\ =\quad \sum _{ k=1 }^{ 13 }{ \frac { \sin { (\frac { \pi }{ 4 } +\frac { k\pi }{ 6 } -[\frac { \pi }{ 4 } +\frac { (k-1)\pi }{ 6 } ]) } }{ \sin { (\frac { \pi }{ 4 } +\frac { k\pi }{ 6 } )\sin { (\frac { \pi }{ 4 } +\frac { (k-1)\pi }{ 6 } ) } } \sin { \frac { \pi }{ 6 } } } } \\ Let\quad A\quad =\quad \frac { \pi }{ 4 } +\frac { k\pi }{ 6 } \quad ;\quad B\quad =\quad \frac { \pi }{ 4 } +\frac { (k-1)\pi }{ 6 } \\ =\quad 2\sum _{ k=1 }^{ 13 }{ \frac { \sin { A\cos { B-\cos { A\sin { B } } } } }{ \sin { A\sin { B } } } } \\ =\quad 2\sum _{ k=1 }^{ 13 }{ \cot { B } } -\cot { A } \\ =\quad 2\quad (\cot { \frac { \pi }{ 4 } } -\cot { \frac { 5\pi }{ 12 } } )\\ =\quad 2(1-(2-\sqrt { 3 } ))\\ =\quad 2(\sqrt { 3 } -1)}\\$