If `4n alpha = pi`, then the value of `tanalpha*tan2alpha*tan3alpha …..tan(2n-2)alphatan(2n-1)alpha` is:

Options

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

Answer

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

\[ \tan \alpha \cdot \tan 2\alpha \cdot \tan 3\alpha \cdots \tan (2n-2)\alpha \cdot \tan (2n-1)\alpha \]

given that \(4n\alpha = \pi\).

### Step 1: Express \(n\) in terms of \(\alpha\)

From the equation \(4n\alpha = \pi\), we can express \(n\) as:

\[ n = \frac{\pi}{4\alpha} \]

### Step 2: Identify the angles in the product

The angles in the product are:

\[ \alpha, 2\alpha, 3\alpha, \ldots, (2n-2)\alpha, (2n-1)\alpha \]

### Step 3: Substitute \(n\) into the angles

Substituting \(n = \frac{\pi}{4\alpha}\), we can find the last angle:

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

### Step 4: Use the identity for tangent

We can pair the terms in the product. Notice that:

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

This means we can pair \(\tan \alpha\) with \(\tan(2n-1)\alpha\), \(\tan 2\alpha\) with \(\tan(2n-2)\alpha\), and so on.

### Step 5: Pair the terms

The product can be expressed as:

\[ (\tan \alpha \cdot \cot \alpha) \cdot (\tan 2\alpha \cdot \cot 2\alpha) \cdots \]

Each pair \(\tan k\alpha \cdot \cot k\alpha = 1\).

### Step 6: Count the pairs

Since we have \(2n\) terms, we will have \(n\) pairs. Thus, the entire product simplifies to:

\[ 1 \cdot 1 \cdots 1 = 1 \]

### Final Answer

Therefore, the value of the product is:

\[ \boxed{1} \]