1. Use principal value ranges

For inverse trigonometric functions:

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

We need to evaluate

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

Since cosine is even,

So,

Now $5 \in [\pi,2\pi]$, and for $x \in [\pi,2\pi]$,

Hence,

3. Evaluate $\sin^{-1}(\sin 6)$

We want the principal value in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

Now,

because

For $x \in \left[\frac{3\pi}{2},2\pi\right]$,

Therefore,

4. Evaluate $\tan^{-1}(\tan 12)$

The principal value of $\tan^{-1}$ lies in

Since tangent has period $\pi$, reduce $12$ by multiples of $\pi$.

We note that

since

so

which is indeed in the principal interval. Hence,

5. Substitute all values

So the expression becomes

Simplify:

6. Match with the options

Given options are:

• A: $3\pi-11$
• B: $4\pi-9$
• C: $4\pi-11$
• D: $3\pi+1$

Our result is

Now check whether it equals any option modulo nothing (since this is a direct value, not periodic after inverse functions):

while

So it does not match option C, and in fact matches none of the given options.

7. Conclusion

The correct value is

Therefore, the stored answer appears to be incorrect.