Step 1: Let \( f(x) = \tan^{-1}(\sin x + \cos x) \).
Since \(\tan^{-1}(x)\) is an increasing function, \(f(x)\) is increasing where \(\sin x + \cos x\) is increasing.
Step 2: Let \( g(x) = \sin x + \cos x \).
Differentiate:
\[
g'(x) = \cos x - \sin x
\]
Set \( g'(x) > 0 \) for increasing:
\[
\cos x - \sin x > 0 \quad \Rightarrow \quad \tan x < 1
\]
Step 3: Solve inequality \( \tan x < 1 \).
\[
\tan x < 1 \Rightarrow x \in \left(-\frac{3\pi}{4}, \frac{\pi}{4} \right) \pmod{\pi}
\]
So in this interval, \( \sin x + \cos x \) is increasing, and hence \( f(x) \) is increasing.