If `theta=pi/(4n)` then the value of `tanthetatan(2theta)tan(3theta)....tan((2n-1)theta)` is

Explanation

To solve the problem, we need to find the value of the product \( \tan(\theta) \tan(2\theta) \tan(3\theta) \ldots \tan((2n-1)\theta) \) given that \( \theta = \frac{\pi}{4n} \).

### Step-by-Step Solution:

1. **Substitute the value of \( \theta \)**:
\[ \theta = \frac{\pi}{4n} \]
This implies:
\[ 2\theta = \frac{\pi}{2n}, \quad 3\theta = \frac{3\pi}{4n}, \quad \ldots, \quad (2n-1)\theta = \frac{(2n-1)\pi}{4n} \]

2. **Evaluate \( \tan(k\theta) \)** for \( k = 1, 2, \ldots, 2n-1 \):
- For \( k = 1 \):
\[ \tan(\theta) = \tan\left(\frac{\pi}{4n}\right) \]
- For \( k = 2 \):
\[ \tan(2\theta) = \tan\left(\frac{\pi}{2n}\right) \]
- For \( k = 3 \):
\[ \tan(3\theta) = \tan\left(\frac{3\pi}{4n}\right) \]
- Continuing this way up to \( k = 2n-1 \):
\[ \tan((2n-1)\theta) = \tan\left(\frac{(2n-1)\pi}{4n}\right) \]

3. **Using the identity for tangent**:
Notice that:
\[ \tan\left(\frac{(2n-1)\pi}{4n}\right) = \cot\left(\frac{\pi}{4} - \frac{(2n-1)\pi}{4n}\right) = \cot\left(\frac{\pi}{4n}\right) \]
This means:
\[ \tan((2n-1)\theta) = \cot(\theta) \]

4. **Relate \( \tan(k\theta) \) and \( \cot(\theta) \)**:
For all \( k \):
- \( \tan(k\theta) = \tan\left(\frac{k\pi}{4n}\right) \) for \( k = 1, 2, \ldots, 2n-1 \)
- The product can be rewritten using the periodic nature of tangent and cotangent functions.

5. **Calculate the product**:
The product becomes:
\[ \tan\left(\frac{\pi}{4n}\right) \tan\left(\frac{2\pi}{4n}\right) \tan\left(\frac{3\pi}{4n}\right) \ldots \tan\left(\frac{(2n-1)\pi}{4n}\right) \]
Each \( \tan(k\theta) \) for \( k = 1, 2, \ldots, 2n-1 \) will yield a value of 1 when evaluated at \( \theta = \frac{\pi}{4n} \).

6. **Final result**:
Thus, the entire product simplifies to:
\[ \tan(\theta) \tan(2\theta) \tan(3\theta) \ldots \tan((2n-1)\theta) = 1 \]

### Conclusion:
The value of \( \tan(\theta) \tan(2\theta) \tan(3\theta) \ldots \tan((2n-1)\theta) \) is \( 1 \).