We aim to show that \( \sin x + \tan x > 2x \) for \( 0 < x < \frac{\pi}{2} \).

### Step 1: Rewrite the inequality
We need to prove:
\[
\sin x + \tan x > 2x.
\]
Recall that \( \tan x = \frac{\sin x}{\cos x} \), so the inequality becomes:
\[
\sin x + \frac{\sin x}{\cos x} > 2x.
\]

### Step 2: Analyze the behavior of each function
- **For \( \sin x \):**
  - \( \sin x \) is increasing on the interval \( 0 < x < \frac{\pi}{2} \), with \( \sin(0) = 0 \) and \( \sin\left(\frac{\pi}{2}\right) = 1 \).
  
- **For \( \tan x \):**
  - \( \tan x \) is also increasing on \( 0 < x < \frac{\pi}{2} \), with \( \tan(0) = 0 \) and \( \tan\left(\frac{\pi}{2}\right) \to \infty \).
  
- **For \( 2x \):**
  - \( 2x \) is a linear function that increases steadily from \( 0 \) to \( \pi \) as \( x \) increases from \( 0 \) to \( \frac{\pi}{2} \).

### Step 3: Check values at the endpoints
- At \( x = 0 \), both sides of the inequality give:
  \[
  \sin(0) + \tan(0) = 0, \quad 2(0) = 0,
  \]
  so the inequality is satisfied.
  
- As \( x \to \frac{\pi}{2} \), the left-hand side \( \sin x + \tan x \to 1 + \infty = \infty \), while the right-hand side \( 2x \) tends to \( \pi \), so for values of \( x \) near \( \frac{\pi}{2} \), the inequality holds because \( \infty > \pi \).

### Step 4: Check the behavior between \( 0 < x < \frac{\pi}{2} \)
To rigorously prove the inequality holds for all \( x \in (0, \frac{\pi}{2}) \), we analyze the derivative of the difference between the two sides:

Define \( f(x) = \sin x + \tan x - 2x \). We want to show that \( f(x) > 0 \) for all \( x \in (0, \frac{\pi}{2}) \).

The derivative of \( f(x) \) is:
\[
f'(x) = \cos x + \sec^2 x - 2.
\]
Since \( \sec^2 x = 1 + \tan^2 x \), this becomes:
\[
f'(x) = \cos x + 1 + \tan^2 x - 2 = \cos x - 1 + \tan^2 x.
\]
For \( 0 < x < \frac{\pi}{2} \), we know \( \cos x < 1 \) and \( \tan^2 x > 0 \), so \( f'(x) > 0 \) in this interval, meaning \( f(x) \) is increasing.

Since \( f(x) = 0 \) at \( x = 0 \) and \( f(x) \to \infty \) as \( x \to \frac{\pi}{2} \), we conclude that \( f(x) > 0 \) for all \( 0 < x < \frac{\pi}{2} \).

Thus, \( \sin x + \tan x > 2x \) for \( 0 < x < \frac{\pi}{2} \).

### Conclusion:
The inequality \( \sin x + \tan x > 2x \) holds true for all \( x \in (0, \frac{\pi}{2}) \).

Would you like more details or further clarification?

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Here are 5 related questions you might find interesting:

1. Can a similar inequality be proven for other trigonometric functions, such as cosine or cotangent?
2. What is the geometric interpretation of \( \sin x + \tan x > 2x \)?
3. How does the inequality change as \( x \to 0 \)?
4. Can the same method be applied to prove inequalities involving higher powers of \( x \)?
5. How would the inequality behave if the interval was extended beyond \( \frac{\pi}{2} \)?

**Tip:** Derivatives are often useful to determine whether a function is increasing or decreasing over an interval.

Show that sinx + tanx > 2x for 0 < x < π/2.

Explore a detailed mathematical proof that sin(x) + tan(x) is greater than 2x for 0 < x < π/2. This problem demonstrates key trigonometric functions, derivative analysis, and function behavior analysis, suitable for advanced high school or undergraduate students.

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