Show that the function $f$ given by $f(x) = \tan^{-1}(\sin x + \cos x), x > 0$ is always an increasing function in $\left(0, \frac{\pi}{4}\right)$.

(N/A) Given $f(x) = \tan^{-1}(\sin x + \cos x)$.
Taking the derivative with respect to $x$:
$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x)$
$f'(x) = \frac{\cos x - \sin x}{1 + (\sin^2 x + \cos^2 x + 2\sin x \cos x)}$
Since $\sin^2 x + \cos^2 x = 1$ and $2\sin x \cos x = \sin 2x$,we have:
$f'(x) = \frac{\cos x - \sin x}{1 + (1 + \sin 2x)} = \frac{\cos x - \sin x}{2 + \sin 2x}$
For $x \in \left(0, \frac{\pi}{4}\right)$,we know that $\cos x > \sin x$,so $\cos x - \sin x > 0$.
Also,$2 + \sin 2x > 0$ for all $x$.
Since both the numerator and denominator are positive in the interval $\left(0, \frac{\pi}{4}\right)$,$f'(x) > 0$.
Therefore,the function $f(x)$ is strictly increasing in $\left(0, \frac{\pi}{4}\right)$.