If `alpha, beta` are two values lying between 0 and `2pi` for which `tan theta = k` then the value of `frac{tan alpha}{2} *frac{tan beta}{2}` is....

Options

A. -1
B. 1
C. `-2/k`
D. none of these

Answer

To solve the problem, we need to find the value of \(\frac{\tan \alpha}{2} \cdot \frac{\tan \beta}{2}\) given that \(\tan \theta = k\) where \(\alpha\) and \(\beta\) are the roots of the equation derived from this tangent function. Here’s a step-by-step solution:

### Step 1: Understand the relationship
Given that \(\tan \theta = k\), we can express this in terms of a quadratic equation. The equation can be written as:
\[ \tan \theta - k = 0 \]
This implies that \(\alpha\) and \(\beta\) are the angles for which this equation holds true.

### Step 2: Use the half-angle formula
We know the half-angle formula for tangent:
\[ \tan \theta = \frac{2 \tan \frac{\theta}{2}}{1 - \tan^2 \frac{\theta}{2}} \]
Substituting \(\theta\) with \(2\alpha\) and \(2\beta\), we get:
\[ \tan 2\alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} \]
\[ \tan 2\beta = \frac{2 \tan \beta}{1 - \tan^2 \beta} \]

### Step 3: Set up the equation
From the equation \(\tan \theta = k\), we can write:
\[ \frac{2 \tan \frac{\alpha}{2}}{1 - \tan^2 \frac{\alpha}{2}} = k \]
Cross-multiplying gives:
\[ 2 \tan \frac{\alpha}{2} = k(1 - \tan^2 \frac{\alpha}{2}) \]
Rearranging gives:
\[ k \tan^2 \frac{\alpha}{2} + 2 \tan \frac{\alpha}{2} - k = 0 \]

### Step 4: Identify the quadratic form
This is a quadratic equation in terms of \(\tan \frac{\alpha}{2}\):
\[ k x^2 + 2x - k = 0 \]
where \(x = \tan \frac{\alpha}{2}\).

### Step 5: Find the product of the roots
For a quadratic equation \(ax^2 + bx + c = 0\), the product of the roots is given by:
\[ \text{Product of roots} = \frac{c}{a} \]
In our case:
\[ \text{Product of roots} = \frac{-(-k)}{k} = \frac{k}{k} = 1 \]
Thus, the product of the roots \(\tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2} = 1\).

### Final Answer
Therefore, the value of \(\frac{\tan \alpha}{2} \cdot \frac{\tan \beta}{2}\) is:
\[ \boxed{1} \]