Range of Tricky Tangent Function

Please educate me if what I did was not correct.

$f\left( x \right) =y=\frac { 1+\tan { x } }{ 1-\tan { x } }$ ­

I took the inverse, $f^{ -1 }\left( x \right)$ , of the function and took its domain as the range of $f\left( x \right)$

$\Rightarrow x=\frac { 1+\tan { y } }{ 1-\tan { y } }$

$\Rightarrow x-x\tan { y } =1+\tan { y }$

$\Rightarrow \tan { y } =\frac { x-1 }{ x+1 }$

$\Rightarrow f^{ -1 }\left( x \right) =\arctan { \frac { x-1 }{ x+1 } } ­$

The denominator tells us that $x\neq -1$ ; $\arctan { }$ can take any value for its argument.

Hence, the Domain of $f^{ -1 }\left( x \right)$ , and Range of $f\left( x \right)$ is

$\left( -\infty ,-1 \right) \cup \left( -1,\infty \right) ­)$ .

Discontinuities happen when $\begin{cases} \displaystyle \lim_{x \to \frac{\pi}{4}^-} \dfrac{1+\tan{x}}{1-\tan{x}} = - \infty \\ \displaystyle \lim_{x \to \frac{\pi}{4}^+} \dfrac{1+\tan{x}}{1-\tan{x}} = + \infty \end{cases}$

Therefore, the range of $f(x)$ is $\boxed{(-\infty, \infty)}$ .

$\Rightarrow \dfrac{1 + \tan x}{1 - \tan x}$ $\Rightarrow \dfrac{ \tan x + \tan\left(\dfrac{\pi}{4}\right)}{1 - \tan x \tan\left(\dfrac{\pi}{4}\right)}$ $\Rightarrow \tan\left(x + \dfrac{\pi}{4}\right)$ $\text{Range of} \tan(x)\ \text{is} \ (-\infty , \infty)$ $\therefore \text{Range of} \tan\left(x + \dfrac{\pi}{4}\right) \ \text{is also} \ (-\infty , \infty)$

f(X)=tan(X+π/4)

Simply take the function as y........
Manipulation leads to
Tan(x) = (y-1)/(y+1)
Thus, y can take any real values except -1........