If `theta_1a n dtheta_2` are two values lying in `[2,2pi]` for which `t a ntheta=lambda,` then `tan(theta_1)/2tan(theta_2)/2` is equal to 0 (b) `-1` (c) 2 (d) 1

Options

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

Explanation

To solve the problem, we need to find the value of \(\frac{\tan(\theta_1/2)}{\tan(\theta_2/2)}\) given that \(\tan(\theta) = \lambda\) for two angles \(\theta_1\) and \(\theta_2\) in the interval \([2, 2\pi]\).

### Step-by-Step Solution:

1. **Start with the given equation**:
We know that:
\[ \tan(\theta) = \lambda \]
This means we can express \(\theta\) in terms of \(\lambda\).

2. **Use the double angle formula for tangent**:
The double angle formula for tangent states:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
Applying this to \(\theta_1\) and \(\theta_2\):
\[ \tan(\theta_1) = \lambda \quad \text{and} \quad \tan(\theta_2) = \lambda \]
Therefore, we can write:
\[ \tan(2\theta_1) = \frac{2\tan(\theta_1)}{1 - \tan^2(\theta_1)} = \frac{2\lambda}{1 - \lambda^2} \]
and similarly for \(\theta_2\):
\[ \tan(2\theta_2) = \frac{2\tan(\theta_2)}{1 - \tan^2(\theta_2)} = \frac{2\lambda}{1 - \lambda^2} \]

3. **Express \(\tan(\theta/2)\)**:
We can rearrange the double angle formula to express \(\tan(\theta/2)\):
\[ \tan(\theta/2) = \frac{\tan(\theta)}{1 + \sqrt{1 + \tan^2(\theta)}} \]
Substituting \(\tan(\theta) = \lambda\):
\[ \tan(\theta/2) = \frac{\lambda}{1 + \sqrt{1 + \lambda^2}} \]

4. **Calculate \(\tan(\theta_1/2)\) and \(\tan(\theta_2/2)\)**:
Using the formula derived above:
\[ \tan(\theta_1/2) = \frac{\lambda}{1 + \sqrt{1 + \lambda^2}} \quad \text{and} \quad \tan(\theta_2/2) = \frac{\lambda}{1 + \sqrt{1 + \lambda^2}} \]

5. **Find the ratio**:
Now we need to find:
\[ \frac{\tan(\theta_1/2)}{\tan(\theta_2/2)} = \frac{\frac{\lambda}{1 + \sqrt{1 + \lambda^2}}}{\frac{\lambda}{1 + \sqrt{1 + \lambda^2}}} = 1 \]

Thus, the final answer is:
\[ \frac{\tan(\theta_1/2)}{\tan(\theta_2/2)} = 1 \]

### Final Answer:
(d) 1