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

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

Differentiating with respect to $x$, we get:

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

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

For $f(x)$ to be an increasing function, $f'(x) > 0$.

Since the denominator $1 + (\sin x + \cos x)^2$ is always positive for all real $x$, the sign of $f'(x)$ depends entirely on the numerator.

$\Rightarrow \cos x - \sin x > 0$

$\Rightarrow \cos x > \sin x$

From the standard trigonometric graphs, the inequality $\cos x > \sin x$ holds true in the interval $\left(\dfrac{-\pi}{2}, \dfrac{\pi}{4}\right)$.

Therefore, the function is increasing in the interval $\left(\dfrac{-\pi}{2}, \dfrac{\pi}{4}\right)$.

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