If `n=pi/(4alpha),` then `tan alpha tan 2alpha tan 3 alpha ... tan(2n-1)alpha` is equal to (a) 1 (b) 1/2 (c) 2 (d) 1/3

Options

A. 1
B. `1//2`
C. 2
D. `1//3`

Explanation

To solve the problem, we need to evaluate the expression \( \tan \alpha \tan 2\alpha \tan 3\alpha \ldots \tan(2n-1)\alpha \) given that \( n = \frac{\pi}{4\alpha} \).

### Step-by-Step Solution:

1. **Identify the Value of \( n \)**:
Given \( n = \frac{\pi}{4\alpha} \), we can express \( 2n \) as:
\[ 2n = 2 \cdot \frac{\pi}{4\alpha} = \frac{\pi}{2\alpha} \]

2. **Determine the Range of Angles**:
The angles we need to evaluate are \( \alpha, 2\alpha, 3\alpha, \ldots, (2n-1)\alpha \). The maximum angle is:
\[ (2n-1)\alpha = \left(2 \cdot \frac{\pi}{4\alpha} - 1\right)\alpha = \left(\frac{\pi}{2\alpha} - 1\right)\alpha \]

3. **Calculate \( 2n - 1 \)**:
We can calculate \( 2n - 1 \):
\[ 2n - 1 = 2 \cdot \frac{\pi}{4\alpha} - 1 = \frac{\pi}{2\alpha} - 1 \]

4. **Evaluate the Product**:
The product we need to evaluate is:
\[ \tan \alpha \tan 2\alpha \tan 3\alpha \ldots \tan(2n-1)\alpha \]
We can pair the terms:
\[ \tan k\alpha \tan((2n-k)\alpha) \text{ for } k = 1, 2, \ldots, n-1 \]
Each pair can be simplified using the identity \( \tan(\frac{\pi}{2} - x) = \cot x \):
\[ \tan k\alpha \tan((2n-k)\alpha) = \tan k\alpha \cot k\alpha = 1 \]

5. **Count the Number of Terms**:
Since \( n = \frac{\pi}{4\alpha} \), the total number of terms is \( n \). The pairing will yield \( n-1 \) pairs, and the middle term when \( n \) is odd will be \( \tan(n\alpha) \) which simplifies to \( 1 \).

6. **Final Calculation**:
Since each pair contributes \( 1 \) and the middle term also contributes \( 1 \), the total product is:
\[ 1 \cdot 1 \cdots 1 = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]