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

For $f(x)$ to be increasing, $f^{\prime }(x)>0$.
$$f^{\prime }(x)=\frac{1}{1+(\sin x+\cos x{)}^2}\cdot \frac{d}{dx}(\sin x+\cos x)$$
$$f^{\prime }(x)=\frac{\cos x-\sin x}{1+(\sin x+\cos x{)}^2}$$
Since the denominator is always positive, we need:
$$\cos x-\sin x>0$$
$$\cos x>\sin x$$
In the region near the origin, $\cos x>\sin x$ when $x<\pi /4$.Considering the interval $(-\pi /2,\pi /4)$, $\cos x$ is positive and greater than $\sin x$ (which is negative in $(-\pi /2,0)$ and positive but smaller than $\cos x$ in $(0,\pi /4)$).Thus, the function is increasing in $(-\pi /2,\pi /4)$.