`If 0 lt alpha lt beta lt (pi)/(2)` then

Options

A. `(tan beta)/(tan alpha ) lt (alpha)/(beta)`
B. `(tan beta)/(tan alpha) gt (alpha)/(beta)`
C. `(tan beta )/(tan alpha ) gt (alpha)/(beta)`
D. `(tan alpha)/(tan beta )le (alpha)/(beta)`

Explanation

To solve the problem, we need to analyze the function \( f(x) = x \tan(x) \) over the interval \( (0, \frac{\pi}{2}) \) and determine the relationship between \( \tan(\alpha) \) and \( \tan(\beta) \) given that \( 0 < \alpha < \beta < \frac{\pi}{2} \).

### Step-by-step Solution:

1. **Define the function**: Let \( f(x) = x \tan(x) \) where \( x \) is in the interval \( (0, \frac{\pi}{2}) \).

2. **Differentiate the function**: We need to find the derivative \( f'(x) \) to analyze the behavior of the function.
\[ f'(x) = \frac{d}{dx}(x \tan(x)) = \tan(x) + x \sec^2(x) \]
Here, we used the product rule for differentiation.

3. **Analyze the derivative**: Since \( \tan(x) \) and \( \sec^2(x) \) are both positive in the interval \( (0, \frac{\pi}{2}) \), we can conclude that:
\[ f'(x) > 0 \quad \text{for } x \in (0, \frac{\pi}{2}) \]
This means that \( f(x) \) is an increasing function in this interval.

4. **Apply the increasing function property**: Given that \( \alpha < \beta \), and since \( f(x) \) is increasing, we have:
\[ f(\alpha) < f(\beta) \]

5. **Substitute the function values**: Now substituting back the function values:
\[ \alpha \tan(\alpha) < \beta \tan(\beta) \]

6. **Rearranging the inequality**: We can rearrange this inequality to compare \( \tan(\alpha) \) and \( \tan(\beta) \):
\[ \frac{\tan(\beta)}{\tan(\alpha)} > \frac{\alpha}{\beta} \]

7. **Conclusion**: Therefore, we conclude that:
\[ \frac{\tan(\beta)}{\tan(\alpha)} > \frac{\alpha}{\beta} \]

### Final Answer:
The correct relationship is:
\[ \tan(\beta) > \frac{\alpha}{\beta} \tan(\alpha) \]