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

We have to compute the expression $$$\cos^{-1}(\cos(-5))+\sin^{-1}(\sin 6)-\tan^{-1}(\tan 12)$$$ where every inverse trigonometric function is to be taken in its principal value branch.

First, recall the principal value ranges:

$$$\cos^{-1}x\in[0,\pi],\qquad \sin^{-1}x\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right],\qquad \tan^{-1}x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right).$$$

1. Evaluating $$\cos^{-1}(\cos(-5))$$

Because the cosine function is even, $$\cos(-5)=\cos 5.$$ However, the principal value of $$\cos^{-1}$$ must lie in $$[0,\pi].$$ Now $$5$$ radians exceeds $$\pi$$ (since $$\pi\approx3.14$$) but is less than $$2\pi$$. Using the identity $$\cos\theta=\cos(2\pi-\theta),$$ we write

$$\cos 5=\cos(2\pi-5).$$

The angle $$2\pi-5$$ clearly lies between $$0$$ and $$\pi$$ because $$0<2\pi-5<\pi.$$ Therefore

$$\cos^{-1}(\cos(-5))=2\pi-5.$$

2. Evaluating $$\sin^{-1}(\sin 6)$$

The angle $$6$$ radians is slightly less than $$2\pi$$, so it is in the interval $$(\pi,2\pi).$$ For such angles the sine is negative and we use the fact that

$$\sin\theta=\sin(\theta-2\pi).$$

Subtracting one full revolution gives

$$6-2\pi\;(\approx -0.283)\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right].$$

Since this result already lies in the principal value range of $$\sin^{-1},$$ we have

$$\sin^{-1}(\sin 6)=6-2\pi.$$

3. Evaluating $$\tan^{-1}(\tan 12)$$

The tangent function has period $$\pi,$$ so we may subtract integer multiples of $$\pi$$ until we land in the interval $$\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right).$$ Write $$12=n\pi+\alpha$$ with $$\alpha\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right).$$

Dividing by $$\pi$$ we find $$\dfrac{12}{\pi}\approx3.82,$$ so choose $$n=4.$$ Then

$$\alpha=12-4\pi.$$

Numerically $$12-4\pi\approx -0.566,$$ which indeed lies between $$-\dfrac{\pi}{2}$$ and $$\dfrac{\pi}{2}.$$ Thus

$$$\tan 12=\tan(12-4\pi)\quad\text{and}\quad\tan^{-1}(\tan 12)=12-4\pi.$$$

4. Combining all three parts

Substituting the three principal values we obtained:

$$$\begin{aligned} \cos^{-1}(\cos(-5))+\sin^{-1}(\sin 6)-\tan^{-1}(\tan 12) &=(2\pi-5)+(6-2\pi)-(12-4\pi)\\ &=2\pi-5+6-2\pi-12+4\pi\\ &=(2\pi-2\pi+4\pi)+(-5+6-12)\\ &=4\pi-11. \end{aligned}$$$

Hence, the correct answer is Option C.