$\text { The function } f(x)=\tan ^{-1}(\sin x+\cos x) \text { is an increasing function in }$

Given $f(x) = \tan^{-1}(\sin x + \cos x)$.

To determine the interval where $f(x)$ is increasing, we find the derivative $f'(x)$ and set $f'(x) > 0$.

$f'(x) = \dfrac{1}{1 + (\sin x + \cos x)^2} \cdot \dfrac{d}{dx}(\sin x + \cos x)$

$f'(x) = \dfrac{\cos x - \sin x}{1 + (\sin^2 x + \cos^2 x + 2 \sin x \cos x)}$

$f'(x) = \dfrac{\cos x - \sin x}{1 + 1 + \sin(2x)} = \dfrac{\cos x - \sin x}{2 + \sin(2x)}$

Since the denominator $2 + \sin(2x)$ is always positive for all real $x$, the sign of $f'(x)$ depends only on the numerator $\cos x - \sin x$.

For $f(x)$ to be increasing, we require $f'(x) > 0$, which implies $\cos x - \sin x > 0$.

$\cos x > \sin x$

Dividing by $\cos x$ (assuming $\cos x > 0$, which holds in the interval $(-\pi/2, \pi/2)$):

$1 > \tan x$

$\tan x < 1$

In the interval $(-\pi/2, \pi/2)$, $\tan x < 1$ holds when $x < \pi/4$.

Thus, the function is increasing in the interval $(-\pi/2, \pi/4)$.

Answer: $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{4}\right)$