Question No. 39

The value of ${\sum }_{k=1}^{13}\frac{1}{\sin (\frac{\pi }{4}+\frac{(k-1)\pi }{6})\sin (\frac{\pi }{4}+\frac{k\pi }{6})}$ is equal to

(A) $3-\sqrt{3}$
(B) $2(3-\sqrt{3})$
(C) $2(\sqrt{3}-1)$
(D) $2(2+\sqrt{3})$

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Solution:

Step 1: General Term

Let the general term be:

$$T_k=\frac{1}{\sin (\frac{\pi }{4}+\frac{(k-1)\pi }{6})\sin (\frac{\pi }{4}+\frac{k\pi }{6})}$$

Step 2: Use Trigonometric Identity

We use the identity:

$$\frac{1}{\sin A\sin B}=\frac{2}{\cos (A-B)-\cos (A+B)}\cdot \frac{1}{2}\cdot 2$$

But a more direct approach is to use the difference of cotangents. Note:

$$\cot A-\cot B=\frac{\sin (B-A)}{\sin A\sin B}$$

So,

$$\frac{1}{\sin A\sin B}=\frac{\cot A-\cot B}{\sin (B-A)}$$

Here, let:

$$A=\frac{\pi }{4}+\frac{(k-1)\pi }{6},B=\frac{\pi }{4}+\frac{k\pi }{6}$$

Then,

$$B-A=\frac{\pi }{6}$$

So,

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

Step 3: Telescoping Series

Let:

$$V_{k-1}=2\cot (\frac{\pi }{4}+\frac{(k-1)\pi }{6}),V_k=2\cot (\frac{\pi }{4}+\frac{k\pi }{6})$$

Then:

$$T_k=V_{k-1}-V_k$$

The sum from $k=1$ to $13$ is:

$$S=\sum _{k=1}^{13}{T}_k=(V_0-V_1)+(V_1-V_2)+\cdots +(V_{12}-V_{13})=V_0-V_{13}$$

So,

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

Step 4: Simplify the Angles

$$\cot \left(\frac{\pi }{4}\right)=1$$$$\frac{\pi }{4}+\frac{13\pi }{6}=\frac{3\pi +26\pi }{12}=\frac{29\pi }{12}$$

But we can reduce this modulo $\pi$:

$$\frac{29\pi }{12}=2\pi +\frac{5\pi }{12}(\text{since }2\pi =\frac{24\pi }{12})$$

So,

$$\cot \left(\frac{29\pi }{12}\right)=\cot \left(\frac{5\pi }{12}\right)$$

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

$$\cot ({75}^{\circ })=\cot ({45}^{\circ }+{30}^{\circ })=\frac{\cot {45}^{\circ }\cot {30}^{\circ }-1}{\cot {45}^{\circ }+\cot {30}^{\circ }}=\frac{1\cdot \sqrt{3}-1}{1+\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$$

Rationalize:

$$=\frac{(\sqrt{3}-1{)}^2}{3-1}=\frac{3-2\sqrt{3}+1}{2}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$$

Step 5: Compute the Sum

$$S=2(1)-2(2-\sqrt{3})=2-4+2\sqrt{3}=2\sqrt{3}-2=2(\sqrt{3}-1)$$

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Final Answer:

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