The value of `sum_(k=1)^(13) (1)/(sin((pi)/(4) + ((k-1)pi)/(6)) sin ((pi)/(4)+ (kpi)/(6)))` is equal to

Options

A. ` 3-sqrt3`
B. `2(3-sqrt3)`
C. `2(sqrt3-1)`
D. `2(2+sqrt3)`

Explanation

To solve the given problem, we need to evaluate the summation:

\[ S = \sum_{k=1}^{13} \frac{1}{\sin\left(\frac{\pi}{4} + \frac{(k-1)\pi}{6}\right) \sin\left(\frac{\pi}{4} + \frac{k\pi}{6}\right)} \]

### Step 1: Rewrite the Expression
We start by rewriting the expression to facilitate simplification. We can multiply and divide by \(\sin\left(\frac{\pi}{6}\right)\):

\[ S = \sum_{k=1}^{13} \frac{\sin\left(\frac{\pi}{6}\right)}{\sin\left(\frac{\pi}{6}\right) \sin\left(\frac{\pi}{4} + \frac{(k-1)\pi}{6}\right) \sin\left(\frac{\pi}{4} + \frac{k\pi}{6}\right)} \]

### Step 2: Use the Identity for Sine
Using the identity \(\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]\), we can express the product of sines in the denominator:

\[ \sin\left(\frac{\pi}{4} + \frac{(k-1)\pi}{6}\right) \sin\left(\frac{\pi}{4} + \frac{k\pi}{6}\right) = \frac{1}{2} \left[\cos\left(\left(\frac{\pi}{4} + \frac{k\pi}{6}\right) - \left(\frac{\pi}{4} + \frac{(k-1)\pi}{6}\right)\right) - \cos\left(\left(\frac{\pi}{4} + \frac{k\pi}{6}\right) + \left(\frac{\pi}{4} + \frac{(k-1)\pi}{6}\right)\right)\right] \]

### Step 3: Simplify the Expression
Now, we can simplify the expression further:

\[ S = \sum_{k=1}^{13} \frac{2 \sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{4} + \frac{k\pi}{6}\right) - \cos\left(\frac{\pi}{4} + \frac{(k-1)\pi}{6}\right)} \]

### Step 4: Use Cotangent Identity
Using the cotangent identity, we can express the sum in terms of cotangent:

\[ S = 2 \sum_{k=1}^{13} \cot\left(\frac{\pi}{4} + \frac{(k-1)\pi}{6}\right) - \cot\left(\frac{\pi}{4} + \frac{k\pi}{6}\right) \]

### Step 5: Evaluate the Summation
This expression is a telescoping series, which means most terms will cancel out:

\[ S = 2 \left( \cot\left(\frac{\pi}{4}\right) - \cot\left(\frac{\pi}{4} + \frac{13\pi}{6}\right) \right) \]

### Step 6: Calculate Cotangent Values
We know that \(\cot\left(\frac{\pi}{4}\right) = 1\) and we need to calculate \(\cot\left(\frac{\pi}{4} + \frac{13\pi}{6}\right)\).

Calculating \(\frac{13\pi}{6}\):

\[ \frac{13\pi}{6} = 2\pi + \frac{\pi}{6} \quad \text{(which is cotangent periodic)} \]

Thus,

\[ \cot\left(\frac{\pi}{4} + \frac{13\pi}{6}\right) = \cot\left(\frac{\pi}{4} + \frac{\pi}{6}\right) = \cot\left(\frac{5\pi}{12}\right) \]

### Step 7: Final Calculation
Now, we know that:

\[ \cot\left(\frac{5\pi}{12}\right) = 2 - \sqrt{3} \]

Thus,

\[ S = 2 \left(1 - (2 - \sqrt{3})\right) = 2\left(\sqrt{3} - 1\right) \]

### Final Answer
The final value of the summation is:

\[ \boxed{2(\sqrt{3} - 1)} \]