Step 1: Find the derivative

Given $f(x) = \cot^{-1}(\sin x + \cos x)$. Using the chain rule, where $\frac{d}{dx}(\cot^{-1} u) = -\frac{1}{1+u^2} \cdot \frac{du}{dx}$:

Step 2: Simplify the expression

Expand the denominator: $1 + (\sin^2 x + \cos^2 x + 2\sin x \cos x) = 1 + (1 + \sin 2x) = 2 + \sin 2x$. Thus:

Step 3: Determine critical points

Set $f'(x) = 0$, which implies $\cos x - \sin x = 0$, so $\tan x = 1$. In the interval $(0, \pi)$, $x = \frac{\pi}{4}$.

Step 4: Analyze intervals

Test the sign of $f'(x)$ in $(0, \frac{\pi}{4})$ and $(\frac{\pi}{4}, \pi)$:

For $x \in (0, \frac{\pi}{4})$, $\cos x > \sin x$, so $f'(x) < 0$ (Decreasing).

For $x \in (\frac{\pi}{4}, \pi)$, $\cos x < \sin x$, so $f'(x) > 0$ (Increasing).