The sum of the absolute maximum and absolute minimum values of the function $$f(x) = \tan^{-1}(\sin x - \cos x)$$ in the interval $$[0, \pi]$$ is

We have $$f(x) = \tan^{-1}(\sin x - \cos x)$$ on $$[0, \pi]$$. Since $$\tan^{-1}$$ is an increasing function, the extrema of $$f(x)$$ correspond to the extrema of $$g(x) = \sin x - \cos x$$.

Now $$g(x) = \sin x - \cos x = \sqrt{2}\sin\left(x - \frac{\pi}{4}\right)$$.

On $$[0, \pi]$$, the argument $$x - \frac{\pi}{4}$$ ranges over $$\left[-\frac{\pi}{4}, \frac{3\pi}{4}\right]$$.

The maximum of $$\sin\left(x - \frac{\pi}{4}\right)$$ on this interval is 1, occurring at $$x - \frac{\pi}{4} = \frac{\pi}{2}$$, i.e., $$x = \frac{3\pi}{4}$$. So the maximum of $$g$$ is $$\sqrt{2}$$.

The minimum of $$\sin\left(x - \frac{\pi}{4}\right)$$ on $$\left[-\frac{\pi}{4}, \frac{3\pi}{4}\right]$$ is $$\sin\left(-\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$, occurring at $$x = 0$$. So the minimum of $$g$$ is $$\sqrt{2} \cdot \left(-\frac{\sqrt{2}}{2}\right) = -1$$.

Therefore:

Absolute maximum of $$f = \tan^{-1}(\sqrt{2})$$

Absolute minimum of $$f = \tan^{-1}(-1) = -\frac{\pi}{4}$$

The sum is $$\tan^{-1}(\sqrt{2}) - \frac{\pi}{4}$$.

Now we need to check if $$\tan^{-1}(\sqrt{2}) = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$$. If $$\theta = \tan^{-1}(\sqrt{2})$$, then $$\tan\theta = \sqrt{2}$$, so $$\sec^2\theta = 1 + 2 = 3$$, giving $$\cos^2\theta = \frac{1}{3}$$, hence $$\cos\theta = \frac{1}{\sqrt{3}}$$ (since $$\theta$$ is in the first quadrant). Therefore $$\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$$.

So the sum equals $$\cos^{-1}\left(\frac{1}{\sqrt{3}}\right) - \frac{\pi}{4}$$.

Hence, the correct answer is Option 3.