1. Let

$$A_r=\frac{\pi}{4}+(r-1)\frac{\pi}{6}.$$

Then

$$A_{r+1}=A_r+\frac{\pi}{6}.$$

So the sum becomes

$$S=\sum_{r=1}^{13}\frac{1}{\sin A_r\sin A_{r+1}}.$$

2. Use the identity

$$\cot x-\cot y=\frac{\sin(y-x)}{\sin x\sin y}.$$

Taking $y=x+\frac{\pi}{6}$,

$$\cot x-\cot\left(x+\frac{\pi}{6}\right) =\frac{\sin\frac{\pi}{6}}{\sin x\sin\left(x+\frac{\pi}{6}\right)} =\frac{1/2}{\sin x\sin\left(x+\frac{\pi}{6}\right)}.$$

Hence,

$$\frac{1}{\sin x\sin\left(x+\frac{\pi}{6}\right)} =2\left[\cot x-\cot\left(x+\frac{\pi}{6}\right)\right].$$

Therefore,

$$S=2\sum_{r=1}^{13}\left(\cot A_r-\cot A_{r+1}\right).$$

This is a telescoping series:

$$S=2(\cot A_1-\cot A_{14}).$$

3. Now compute $A_1$ and $A_{14}$:

$$A_1=\frac{\pi}{4},$$

and

$$A_{14}=\frac{\pi}{4}+13\cdot\frac{\pi}{6} =\frac{3\pi+26\pi}{12}=\frac{29\pi}{12}.$$

Since cotangent has period $\pi$,

$$\cot\frac{29\pi}{12}=\cot\frac{5\pi}{12}.$$

Thus,

$$S=2\left(\cot\frac{\pi}{4}-\cot\frac{5\pi}{12}\right).$$

4. Evaluate the cotangents:

$$\cot\frac{\pi}{4}=1.$$

Also,

$$\frac{5\pi}{12}=75^\circ,$$

so

$$\tan 75^\circ=2+\sqrt{3} \quad\Rightarrow\quad \cot 75^\circ=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}.$$

Hence,

$$S=2\left(1-(2-\sqrt{3})\right)=2(\sqrt{3}-1)=2\sqrt{3}-2.$$

5. Compare with

$$S=a\sqrt{3}+b.$$

So,

$$a=2,\qquad b=-2.$$

Therefore,

$$a^2+b^2=2^2+(-2)^2=4+4=8.$$

6. Option check:

• A: $10$ ✗
• B: $4$ ✗
• C: $8$ ✓
• D: $2$ ✗

So the correct option is C.