How do you evaluate $\arccos (1)$?

Hint: We already know that tan or tangent function is well-defined and one to one. So its inverse exists and its inverse is called arccosecant or shortly as $\arccos $. Thus the domain of $arccos$ is the range of cos which is the set of numbers $(-1,1)$ and its range is equal to the domain of the cosecant function, that is the set $[0,\pi ]$. So, we can evaluate the value of arctan easily if we know the value of tan. Let us suppose $\cos x=y$.
Then $\arccos (y)={{\cos }^{-1}}(y)={{\cos }^{-1}}(\cos (x))$ =x

Hence this way is the most suitable to find the value of arccos. Although there exists a complete list of values of the functions for various values of the domain which could be used to determine the value of Trigonometric functions.

Complete Step by Step Solution:
We have to evaluate the value of $\arccos (1)$
We already know that the value of $\cos 0$ is $1$
Then $\arccos (1)={{\cos }^{-1}}(1)={{\cos }^{-1}}(\cos (0))=0$
Also the value of $\cos 2\pi =1$
thus $\arccos (1)=2\pi $
Infact , in general , $\arccos (1)=2n\pi $ , where $n=0,1,2,...$
But , the range of arcos is $[0,\pi ]$.

Therefore, we have calculated the value of $\arccos (1)$ which is $0$.

Note:
Arctan, arcsin, and arccos functions return angles either in degree or radians as the answer. It is because these angles are the same whose tan, sin, and cos respectively would give the given number.