The function `f(x)=tan^(-1)(sinx+cosx)` is an increasing function in

Options

A. `(-pi/2, 0)`
B. `(0, pi/2)`
C. `(-pi/4, pi/4)`
D. none of these

Answer

To determine the intervals where the function \( f(x) = \tan^{-1}(\sin x + \cos x) \) is increasing, we need to find the derivative \( f'(x) \) and analyze its sign.

### Step 1: Find the derivative of \( f(x) \)

Using the chain rule, we know that the derivative of \( \tan^{-1}(u) \) is given by:

\[ f'(x) = \frac{1}{1 + u^2} \cdot u' \]

where \( u = \sin x + \cos x \).

Now, we differentiate \( u \):

\[ u' = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x \]

Thus, we can express \( f'(x) \) as:

\[ f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x) \]

### Step 2: Determine when \( f'(x) \geq 0 \)

For \( f(x) \) to be an increasing function, we need:

\[ f'(x) \geq 0 \]

This implies:

\[ \cos x - \sin x \geq 0 \]

or equivalently:

\[ \cos x \geq \sin x \]

### Step 3: Solve the inequality \( \cos x \geq \sin x \)

To solve \( \cos x \geq \sin x \), we can rearrange it to:

\[ \cos x - \sin x \geq 0 \]

This can be rewritten as:

\[ \frac{\cos x}{\sin x} \geq 1 \]

or

\[ \tan x \leq 1 \]

### Step 4: Find the intervals

The inequality \( \tan x \leq 1 \) holds true in the intervals where the angle \( x \) is between:

\[ -\frac{\pi}{4} + n\pi \quad \text{and} \quad \frac{\pi}{4} + n\pi \]

for any integer \( n \).

### Step 5: Identify the specific intervals in the range \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \)

Within the interval \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \):

- \( \tan x = 1 \) at \( x = \frac{\pi}{4} \)
- \( \tan x < 1 \) for \( x < \frac{\pi}{4} \)
- \( \tan x > 1 \) for \( x > \frac{\pi}{4} \)

Thus, the function \( f(x) \) is increasing in the interval:

\[ \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \]

### Conclusion

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

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