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(C) We need to evaluate the expression $\cos ^{-1}(\cos (-5))+\sin ^{-1}(\sin (6))-	an ^{-1}(	an (12))$.$1$. For $\cos ^{-1}(\cos (-5))$: Since $\co

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$4 \pi-11$

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(C) We need to evaluate the expression $\cos ^{-1}(\cos (-5))+\sin ^{-1}(\sin (6))-	an ^{-1}(	an (12))$.$1$. For $\cos ^{-1}(\cos (-5))$: Since $\cos(-x) = \cos(x)$,this is $\cos ^{-1}(\cos (5))$. Since $5 \in [\pi, 2\pi]$,we use $\cos ^{-1}(\cos x) = 2\pi - x$. Thus,$\cos ^{-1}(\cos (5)) = 2\pi - 5$.$2$. For $\sin ^{-1}(\sin (6))$: Since $6 \in [\frac{3\pi}{2}, 2\pi]$,we use $\sin ^{-1}(\sin x) = x - 2\pi$. Thus,$\sin ^{-1}(\sin (6)) = 6 - 2\pi$.$3$. For $	an ^{-1}(	an (12))$: Since $12 \in (3\pi + \frac{\pi}{2}, 4\pi + \frac{\pi}{2})$,we use $	an ^{-1}(	an x) = x - 4\pi$. Thus,$	an ^{-1}(	an (12)) = 12 - 4\pi$.Combining these: $(2\pi - 5) + (6 - 2\pi) - (12 - 4\pi) = 2\pi - 5 + 6 - 2\pi - 12 + 4\pi = 4\pi - 11$.

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

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