Step 1: Split the function into two parts:
Let $h(x) = \sin x + \cos x$, so $f(x) = \cot^{-1}(h(x))$.
The function $\cot^{-1}$ is strictly decreasing on all real numbers.
So $f$ does the opposite of $h$: when $h$ rises, $f$ falls, and when $h$ falls, $f$ rises.
Step 2: Rewrite h in a single sine:
\[ h(x) = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right) \]
For $x \in (0,\pi)$, the angle $x + \frac{\pi}{4}$ runs over $\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$.
Step 3: Study the sine on that range:
The sine increases up to $\frac{\pi}{2}$ and decreases after it.
The angle equals $\frac{\pi}{2}$ when $x = \frac{\pi}{4}$.
So $h$ is increasing on $\left(0, \frac{\pi}{4}\right)$ and decreasing on $\left(\frac{\pi}{4}, \pi\right)$.
Step 4: Reverse the behaviour for f:
On $\left(0, \frac{\pi}{4}\right)$, $h$ is increasing, so $f$ is decreasing.
On $\left(\frac{\pi}{4}, \pi\right)$, $h$ is decreasing, so $f$ is increasing.
Step 5: Confirm with test points:
The derivative is $f^{\prime}(x) = \dfrac{\sin x - \cos x}{2 + \sin 2x}$.
At $x = \frac{\pi}{6}$, the numerator is $\frac{1}{2} - \frac{\sqrt{3}}{2} < 0$, so $f^{\prime} < 0$.
At $x = \frac{\pi}{2}$, the numerator is $1 - 0 = 1 > 0$ and the denominator is $2$, so $f^{\prime} = \frac{1}{2} > 0$.
Both checks agree with Step 4.
Final Answer:
$f$ decreases on $\left(0, \frac{\pi}{4}\right)$ and increases on $\left(\frac{\pi}{4}, \pi\right)$.\[ \boxed{\text{Decreasing: } \left(0,\frac{\pi}{4}\right), \quad \text{Increasing: } \left(\frac{\pi}{4},\pi\right)} \]