${\cos }^{-1}\left(\cos \left(-5\right)\right)+{\sin }^{-1}\left(\sin \left(6\right)\right)-{\tan }^{-1}\left(\tan \left(12\right)\right)$ is equal to:(The inverse trigonometric functions take the principal values)

$$\begin{gathered} {\cos }^{-1}(\cos 5)+{\sin }^{-1}(\sin 6)-{\tan }^{-1}(\tan 12) \\ {1}^{\mathrm{c}}=57·2 5^{\mathrm{c}}=286=2\mathrm{\pi }-74 \\ {\cos }^{-1}\left(\cos \left(2\mathrm{\pi }-74\right)\right)=74=2\mathrm{\pi }-5 \\ {6}^{\mathrm{c}}=343·2=2\mathrm{\pi }-16÷8 \\ {\sin }^{-1}\left(\sin \right(2\mathrm{\pi }-16·8\left)\right)=-16·8=6-2\mathrm{\pi } \\ {12}^{\mathrm{c}}=686·4=720-33·6=4\mathrm{\pi }-33·6 \\ {\tan }^{-1}\left(\tan \right(4\mathrm{\pi }-33·6\left)\right)=-33·6=12-4\mathrm{\pi } \\ 2\mathrm{\pi }-5+6-2\mathrm{\pi }-12+4\mathrm{\pi }=4\mathrm{\pi }-11 \end{gathered}$$