If $$\sum_{r=1}^{13}\left\{\frac{1}{\sin(\frac{\pi}{4}+(r-1)\frac{\pi}{6})\sin(\frac{\pi}{4}+\frac{r\pi}{6})}\right\}=a\sqrt{3}+b,a,b \in Z$$ then $$a^{2}+b^{2}$$ is equal to:

We have to evaluate the finite sum

$$S=\displaystyle\sum_{r=1}^{13}\frac{1}{\sin\Bigl(\tfrac{\pi}{4}+(r-1)\tfrac{\pi}{6}\Bigr)\; \sin\Bigl(\tfrac{\pi}{4}+r\tfrac{\pi}{6}\Bigr)}.$$

Put $$A_r=\tfrac{\pi}{4}+(r-1)\tfrac{\pi}{6}\qquad (r=1,2,\dots,13).$$
Then the general term of the series can be rewritten as

$$\frac{1}{\sin A_r\;\sin(A_r+\tfrac{\pi}{6})}.$$

Step 1  : Express the reciprocal product through cotangents

The identity $$\cot x-\cot y=\dfrac{\sin(y-x)}{\sin x\;\sin y}$$ holds for all $$x,y$$ (provided the sines are non-zero). Putting $$x=A_r,\;y=A_r+\tfrac{\pi}{6}$$ gives

$$\frac{1}{\sin A_r\;\sin(A_r+\tfrac{\pi}{6})}= \frac{\cot A_r-\cot(A_r+\tfrac{\pi}{6})}{\sin\bigl((A_r+\tfrac{\pi}{6})-A_r\bigr)} =\frac{\cot A_r-\cot(A_r+\tfrac{\pi}{6})}{\sin(\tfrac{\pi}{6})}.$$

Since $$\sin(\tfrac{\pi}{6})=\tfrac12,$$ we obtain the compact form

$$\frac{1}{\sin A_r\;\sin(A_r+\tfrac{\pi}{6})}=2\bigl[\cot A_r-\cot(A_r+\tfrac{\pi}{6})\bigr].$$

Step 2  : Substitute and observe telescoping

Using the above identity, the required sum becomes

$$\begin{aligned} S&=2\sum_{r=1}^{13}\Bigl[\cot A_r-\cot(A_r+\tfrac{\pi}{6})\Bigr]\\ &=2\Bigl[(\cot A_1-\cot A_2)+(\cot A_2-\cot A_3)+\dots+(\cot A_{13}-\cot A_{14})\Bigr]. \end{aligned}$$

All the interior cotangents cancel in pairs, leaving only the first and the last terms:

$$S=2\bigl[\cot A_1-\cot A_{14}\bigr].$$

Step 3  : Evaluate the boundary cotangents

Compute the angles:

$$A_1=\tfrac{\pi}{4},\qquad A_{14}=A_1+13\cdot\tfrac{\pi}{6}=\tfrac{\pi}{4}+\tfrac{13\pi}{6} =\tfrac{29\pi}{12}.$$

The cotangents are

$$\cot A_1=\cot\!\bigl(\tfrac{\pi}{4}\bigr)=1,$$

and, using the $$\pi$$-periodicity of cotangent,

$$\cot A_{14}=\cot\!\bigl(\tfrac{29\pi}{12}\bigr) =\cot\!\bigl(\tfrac{29\pi}{12}-2\pi\bigr) =\cot\!\bigl(\tfrac{5\pi}{12}\bigr)=\cot 75^{\circ}.$$

Because $$\cot 75^{\circ}=\tan 15^{\circ}$$ and $$\tan 15^{\circ}=2-\sqrt3,$$ we have

$$\cot A_{14}=2-\sqrt3.$$

Step 4  : Obtain the numerical value of $$S$$

$$\begin{aligned} S&=2\bigl[\;1-(2-\sqrt3)\bigr]\\ &=2\bigl[-1+\sqrt3\bigr]\\ &=2\sqrt3-2. \end{aligned}$$

Step 5  : Identify $$a$$ and $$b$$

The sum is of the form $$a\sqrt3+b$$ with integers $$a=2,\;b=-2.$$ Therefore

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

Hence the required value is $$\boxed{8}$$, which corresponds to Option D.