Answer
$ a.\quad$
Domain: all reals, except $\displaystyle \frac{\pi}{2}+k\pi,k\in \mathbb{Z}$
Range: $(-\displaystyle \frac{\pi}{2},\frac{\pi}{2})$
$ b.\quad$
Domain: all reals, $(-\infty, \infty)$
Range: $(-\infty, \infty)$
(Additional discussion below)
Graphs:
.
Work Step by Step
$\tan^{-1}x$ is defined for all real numbers
The range is the interval $(-\displaystyle \frac{\pi}{2},\frac{\pi}{2})$
$\tan x$ is defined for all reals, except $\displaystyle \frac{\pi}{2}+k\pi,k\in \mathbb{Z}$
The range of $\tan x$ is $(-\infty, \infty)$
Note that $\tan^{-1}x$ is the inverse of $\tan x$ ONLY after we reduced the domain of tan to $(-\displaystyle \frac{\pi}{2},\frac{\pi}{2})$.
$ a.\quad$
$y=\tan^{-1}(\tan x)$
Domain: all reals, except $\displaystyle \frac{\pi}{2}+k\pi,k\in \mathbb{Z}$
Range: $(-\displaystyle \frac{\pi}{2},\frac{\pi}{2})$
The graph makes sense.
This is not an identity function, because tan was not one-to-one one its whole domain, so we reduced it to $(-\displaystyle \frac{\pi}{2},\frac{\pi}{2})$, where the inverse can be defined.
This means that $\tan^{-1}(\tan 200\pi)$ is not going to be$ 200\pi,$ because
$\tan 200\pi=0$
the number in $(-\displaystyle \frac{\pi}{2},\frac{\pi}{2})$ whose tan equals 0, is the number 0. Thus:
$\tan^{-1}(\tan 200\pi)=0$
$ b.\quad$
$y=\tan(\tan^{-1}x)$
The graph is equal to the graph of $y=x.$
Domain: all reals, $(-\infty, \infty)$
Range: $(-\infty, \infty)$
The graph makes sense.
Take any number $x$. We see that $ \tan^{-1}x=A$ is a value from $(-\displaystyle \frac{\pi}{2},\frac{\pi}{2})$, such that $\tan A=x.$
So $\tan(\tan^{-1}x)=x$
