If `f(x) =sin x +log_(e)|sec x + tanx|-2x for x in (-(pi)/(2),(pi)/(2))` then check the monotonicity of f(x)

Explanation

To determine the monotonicity of the function \( f(x) = \sin x + \log_e |\sec x + \tan x| - 2x \) for \( x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), we will follow these steps:

### Step 1: Differentiate the function
First, we need to find the derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\log_e |\sec x + \tan x|) - \frac{d}{dx}(2x) \]

Calculating each derivative:

- The derivative of \( \sin x \) is \( \cos x \).
- For \( \log_e |\sec x + \tan x| \), we use the chain rule:
\[ \frac{d}{dx}(\log_e u) = \frac{1}{u} \cdot \frac{du}{dx} \]
where \( u = \sec x + \tan x \). The derivatives are:
\[ \frac{du}{dx} = \sec x \tan x + \sec^2 x \]
Therefore,
\[ \frac{d}{dx}(\log_e |\sec x + \tan x|) = \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} \]
- The derivative of \( -2x \) is \( -2 \).

Putting it all together, we have:
\[ f'(x) = \cos x + \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} - 2 \]

### Step 2: Simplify the derivative
We can rewrite \( f'(x) \):
\[ f'(x) = \cos x - 2 + \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} \]

Now, since \( \sec x = \frac{1}{\cos x} \), we can express everything in terms of \( \cos x \):
\[ f'(x) = \cos x - 2 + \frac{\frac{1}{\cos x} \left(\frac{\sin x}{\cos x} + \frac{1}{\cos x}\right)}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} \]
This simplifies to:
\[ f'(x) = \cos x - 2 + \frac{\sin x + 1}{1 + \sin x} \]

### Step 3: Analyze the sign of the derivative
To check the monotonicity, we need to analyze the sign of \( f'(x) \).

1. **Find critical points**: Set \( f'(x) = 0 \):
\[ \cos x - 2 + \frac{\sin x + 1}{1 + \sin x} = 0 \]
This equation is complex, but we can check specific values.

2. **Evaluate at \( x = 0 \)**:
\[ f'(0) = \cos(0) - 2 + \frac{\sin(0) + 1}{1 + \sin(0)} = 1 - 2 + 1 = 0 \]

3. **Check intervals**:
- For \( x < 0 \) (e.g., \( x = -\frac{\pi}{4} \)):
\[ f'(-\frac{\pi}{4}) > 0 \]
- For \( x > 0 \) (e.g., \( x = \frac{\pi}{4} \)):
\[ f'(\frac{\pi}{4}) > 0 \]

### Conclusion
Since \( f'(x) \) is positive for \( x < 0 \) and \( x > 0 \), and \( f'(0) = 0 \), we conclude that \( f(x) \) is strictly increasing in the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).