The function` f(x) = tan^(-1) (sin x + cos x)` is an increasing function in :

Options

A. `( pi/4,(pi)/(2))`
B. `(-(pi)/(2),(pi)/(4))`
C. `(0,(pi)/(2))`
D. `(-(pi)/(2), (pi)/(2))`

Answer

To determine the intervals where the function \( f(x) = \tan^{-1}(\sin x + \cos x) \) is increasing, we need to analyze the derivative of the function, \( f'(x) \), and find where it is greater than or equal to zero.

### Step-by-Step Solution:

1. **Find the Derivative of the Function**:
The derivative of \( f(x) \) is given by the chain rule:
\[ f'(x) = \frac{d}{dx} \left( \tan^{-1}(\sin x + \cos x) \right) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot \frac{d}{dx}(\sin x + \cos x) \]
Now, we differentiate \( \sin x + \cos x \):
\[ \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x \]
Therefore, we have:
\[ f'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2} \]

2. **Determine Where the Derivative is Non-negative**:
For \( f(x) \) to be increasing, we need:
\[ f'(x) \geq 0 \implies \cos x - \sin x \geq 0 \]
This simplifies to:
\[ \cos x \geq \sin x \]

3. **Solve the Inequality**:
The inequality \( \cos x \geq \sin x \) can be rewritten as:
\[ \frac{\cos x}{\sin x} \geq 1 \implies \cot x \geq 1 \]
This means that:
\[ x \leq \frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]

4. **Identify the Relevant Intervals**:
We need to find the intervals where this inequality holds true. The first interval of interest is:
\[ x \in \left[-\frac{\pi}{2}, \frac{\pi}{4}\right] \]
In this interval, \( \cos x \) is greater than or equal to \( \sin x \).

5. **Check Other Intervals**:
- For \( x \in \left(0, \frac{\pi}{4}\right) \): Here, \( \cos x > \sin x \).
- For \( x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \): Here, \( \sin x > \cos x \), so the function is not increasing.

6. **Conclusion**:
The function \( f(x) = \tan^{-1}(\sin x + \cos x) \) is increasing in the interval:
\[ x \in \left[-\frac{\pi}{2}, \frac{\pi}{4}\right] \]