If `4nalpha =pi` then `cot alpha cot 2 alpha cot 3alpha ...cot (2n-1)alpha` `n in Z` is equal to

Options

A. 0
B. 1
C. n
D. None of these

Answer

To solve the problem, we need to find the value of the product \( \cot \alpha \cdot \cot 2\alpha \cdot \cot 3\alpha \cdots \cot (2n-1)\alpha \) given that \( 4n\alpha = \pi \).

### Step-by-Step Solution:

1. **Express \( \alpha \) in terms of \( n \)**:
\[ 4n\alpha = \pi \implies \alpha = \frac{\pi}{4n} \]

**Hint**: Use the equation provided to isolate \( \alpha \).

2. **Calculate the angles for the cotangent terms**:
- The angles we need to evaluate are:
\[ \cot \alpha, \cot 2\alpha, \cot 3\alpha, \ldots, \cot (2n-1)\alpha \]
- Substitute \( \alpha \):
\[ \cot \left( \frac{\pi}{4n} \right), \cot \left( \frac{2\pi}{4n} \right), \cot \left( \frac{3\pi}{4n} \right), \ldots, \cot \left( \frac{(2n-1)\pi}{4n} \right) \]

3. **Simplify the last term**:
- The last term can be simplified:
\[ (2n-1)\alpha = (2n-1) \cdot \frac{\pi}{4n} = \frac{(2n-1)\pi}{4n} \]
- This can be rewritten as:
\[ \frac{\pi}{2} - \frac{\pi}{4n} \]

4. **Use the cotangent identity**:
- Recall the identity:
\[ \cot \left( \frac{\pi}{2} - x \right) = \tan x \]
- Thus, we have:
\[ \cot \left( \frac{(2n-1)\pi}{4n} \right) = \tan \left( \frac{\pi}{4n} \right) \]

5. **Pair the cotangent and tangent terms**:
- The product can be grouped:
\[ \cot \left( \frac{\pi}{4n} \right) \cdot \tan \left( \frac{\pi}{4n} \right) = 1 \]
- Similarly, for each pair:
\[ \cot \left( \frac{2\pi}{4n} \right) \cdot \tan \left( \frac{3\pi}{4n} \right) = 1 \]
- This pattern continues for all terms.

6. **Count the pairs**:
- There are \( n \) pairs in total, and each pair contributes \( 1 \) to the product:
\[ \text{Total product} = 1 \cdot 1 \cdots \text{(n times)} = 1 \]

### Final Result:
The value of \( \cot \alpha \cdot \cot 2\alpha \cdots \cot (2n-1)\alpha \) is:
\[ \boxed{1} \]