If `theta = ( pi )/( 4n )` then the value of `tan theta tan 2 theta "……" tan ( 2 n -2 ) theta tan ( 2 n -1) theta` is

Options

A. `-1`
B. 1
C. 0
D. 2

Answer

To solve the problem, we need to find the value of the expression:

\[ \tan \theta \tan 2\theta \tan 3\theta \ldots \tan (2n-2)\theta \tan (2n-1)\theta \]

where \(\theta = \frac{\pi}{4n}\).

### Step 1: Substitute the value of \(\theta\)
We start by substituting \(\theta\) into the expression:

\[ \tan \left(\frac{\pi}{4n}\right) \tan \left(2 \cdot \frac{\pi}{4n}\right) \tan \left(3 \cdot \frac{\pi}{4n}\right) \ldots \tan \left((2n-2) \cdot \frac{\pi}{4n}\right) \tan \left((2n-1) \cdot \frac{\pi}{4n}\right) \]

This simplifies to:

\[ \tan \left(\frac{\pi}{4n}\right) \tan \left(\frac{\pi}{2n}\right) \tan \left(\frac{3\pi}{4n}\right) \ldots \tan \left(\frac{(2n-2)\pi}{4n}\right) \tan \left(\frac{(2n-1)\pi}{4n}\right) \]

### Step 2: Identify pairs of angles
Notice that for each term \(\tan k\theta\), we can pair it with \(\tan((2n-k)\theta)\):

\[ \tan \left(\frac{k\pi}{4n}\right) \tan \left(\frac{(2n-k)\pi}{4n}\right) \]

For \(k = 1\) to \(n-1\), we can see that:

\[ \tan \left(\frac{k\pi}{4n}\right) \tan \left(\frac{(2n-k)\pi}{4n}\right) = \tan \left(\frac{k\pi}{4n}\right) \tan \left(\frac{\pi}{2} - \frac{k\pi}{4n}\right) = \tan \left(\frac{k\pi}{4n}\right) \cot \left(\frac{k\pi}{4n}\right) = 1 \]

### Step 3: Count the pairs
The pairs will be:

- \((\tan \left(\frac{\pi}{4n}\right), \tan \left(\frac{(2n-1)\pi}{4n}\right))\)
- \((\tan \left(\frac{2\pi}{4n}\right), \tan \left(\frac{(2n-2)\pi}{4n}\right))\)
- ...
- Up to \((\tan \left(\frac{(n-1)\pi}{4n}\right), \tan \left(\frac{(n+1)\pi}{4n}\right))\)

### Step 4: Final value
Since each pair multiplies to 1, and there are \(n-1\) such pairs, the entire product simplifies to:

\[ 1 \cdot 1 \cdot \ldots \cdot 1 = 1 \]

Thus, the value of the expression is:

\[ \boxed{1} \]