The domain of the function $\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\left|{\sin }^{-1}(\sin \mathrm{x})\right|-{\cos }^{-1}(\cos \mathrm{x})}$ in $\left[0,2\mathrm{\pi }\right]$ is 

As, $\left|{\sin }^{-1}(\sin \mathrm{x})\right|$ could be sketched, as

and ${\cos }^{-1}(\cos \mathrm{x})$could be sketched as 

$\therefore \left|{\sin }^{-1}(\sin \mathrm{x})\right|>{\cos }^{-1}(\cos \mathrm{x})$is not possible.
Only equality holds as 

Thus, $\left|{\sin }^{-1}\left(\sin \mathrm{x}\right)\right|={\cos }^{-1}\left(\cos \mathrm{x}\right)$
When                  $\mathrm{x}\in \left[0,\frac{\mathrm{\pi }}{2}\right]\cup \left[\frac{3\mathrm{\pi }}{2},2\mathrm{\pi }\right]$
So, domain for $\mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \left[0,\frac{\mathrm{\pi }}{2}\right]\cup \left[\frac{3\mathrm{\pi }}{2},2\mathrm{\pi }\right]$
Hence, (a) is the correct answer.