If `x_(1),x_(2) in (0,(pi)/(2))`, then `(tan_(x_(2)))/(tanx_(1))` is (where `x_(1)lt x_(2)`)

Options

A. `lt (x_(1))/(x_(2))`
B. `=(x_(1))/(x_(2))`
C. `ltx_(1) x_(2)`
D. `gt(x_(2))/(x_(1))`

Explanation

To solve the problem, we need to analyze the function \( \tan(x) \) over the interval \( (0, \frac{\pi}{2}) \) and its behavior with respect to the given conditions.

### Step-by-Step Solution:

1. **Understanding the Interval**:
We know that \( x_1, x_2 \in (0, \frac{\pi}{2}) \) and \( x_1 < x_2 \). This means both \( x_1 \) and \( x_2 \) are positive angles less than \( \frac{\pi}{2} \).

2. **Behavior of the Tangent Function**:
The function \( \tan(x) \) is known to be an increasing function in the interval \( (0, \frac{\pi}{2}) \). This means that if \( x_1 < x_2 \), then \( \tan(x_1) < \tan(x_2) \).

3. **Setting Up the Ratio**:
We need to analyze the ratio \( \frac{\tan(x_2)}{\tan(x_1)} \). Since \( \tan(x) \) is increasing, we can conclude:
\[ \tan(x_2) > \tan(x_1) \]

4. **Conclusion About the Ratio**:
Since \( \tan(x_2) > \tan(x_1) \), it follows that:
\[ \frac{\tan(x_2)}{\tan(x_1)} > 1 \]

5. **Final Result**:
Therefore, we conclude that:
\[ \frac{\tan(x_2)}{\tan(x_1)} > 1 \]

### Final Answer:
The expression \( \frac{\tan(x_2)}{\tan(x_1)} \) is greater than 1.

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