Question

The range of $f\left(x\right)=\frac{1-\tan x}{1+\tan x}$ is

• A. $(-\infty ,\infty )$
• B. $(-\infty ,0)$
• C. $(0,\infty )$
• D. $(-\infty ,-1)\cup (-1,\infty )$

Answer 1

The correct option is D $(-\infty ,-1)\cup (-1,\infty )$

$\frac{1-\tan x}{1+\tan x}=\frac{y}{1}$
Using componendo, dividendo, we get
$\frac{y-1}{y+1}=\frac{-2\tan x}{2}$
$\tan x=\frac{1-y}{1+y}$
Since the range of $\tan x$ is $(-\infty ,\infty )$ and $y+1\ne 0$
So the range of the above function will be
$(-\infty ,-1)\cup (-1,\infty )$