$\sum_{k=0}^{12} \frac{1}{\sin \left((k+1) \frac{\pi}{6}+\frac{\pi}{4}\right) \sin \left(\frac{k \pi}{6}+\frac{\pi}{4}\right)}=$

• A

  $2(\sqrt{3}+1)$
• B

  $2(3-\sqrt{3})$
• C

  $2(2-\sqrt{3})$
• Correct answer
  D

  $2(\sqrt{3}-1)$

Hence, the correct answer is option: 'D'.