Given the function tan x with domain $\left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$ find the domain of its inverse.
A) $\left[ - \frac { \sqrt { 3 } } { 2 } , \frac { \sqrt { 3 } } { 2 } \right]$
B)[0, $\infty$ )
C)[ $- \pi , \pi$ ]
D)[ $- 1,1$ ]
E) $\left[ - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right]$
F) $\left[ - \frac { 1 } { 2 } , \frac { 1 } { 2 } \right]$
G)( $- \infty , \infty )$
H) $\left[ - \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } \right]$

Question

Given the function tan x with domain $\left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$ find the domain of its inverse.
A) $\left[ - \frac { \sqrt { 3 } } { 2 } , \frac { \sqrt { 3 } } { 2 } \right]$ 
B)[0, $\infty$ )
C)[ $- \pi , \pi$ ]
D)[ $- 1,1$ ]
E) $\left[ - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right]$ 
F) $\left[ - \frac { 1 } { 2 } , \frac { 1 } { 2 } \right]$ 
G)( $- \infty , \infty )$ 
H) $\left[ - \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } \right]$

Quizplus · Accepted Answer

The inverse of the tangent function, $\tan^{-1}(x)$ or arctan(x), has a domain of all real numbers, $(-\infty, \infty)$, because tangent values can range from negative infinity to positive infinity within its original domain of $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Given the Function Tan X with Domain (−π2,π2)\left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)(−2π​,2π​)

Given the Function Tan X with Domain $\left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$