Domain and range of a composite trig function

Question

For $f(x) = \csc(\tan^{-1}x)$ why is the domain $(-\infty, 0)\cup(0,\infty)$ when the range of $\tan^{-1}x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$? I thought the range of the inner function is the domain of the outer function and thus the entire function overall. Also, how would I find the range of $f(x)$?

Answer 1

The implied domain of the composite function $\csc(\arctan x)$ is the largest subset of the domain of $\arctan x$ that maps to elements in the domain of $\csc x$.

$$\arctan x: (-\infty, \infty) \to \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$

Since $$\csc x = \frac{1}{\sin x}$$ the function defined by $\csc x$ is defined for every real number $x$ except those where $\sin x = 0$. Since $\sin x = 0$ when $x = n\pi, n \in \mathbb{Z}$, the only point in the range of $\arctan x$ where $\csc y$ is not defined is $y = 0$.

Since the only number $y$ in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\tan y = 0$ is $y = 0$, $0$ is the only real number in the domain of $\arctan x$ such that $\csc(\arctan x)$ is undefined. Therefore, the implied domain of the function defined by $\csc(\arctan x)$ is $(-\infty, 0) \cup (0, \infty)$.

Since $\csc(\arctan x)$ is continuous on $(0, \infty)$, \begin{align*} \lim_{x \to 0^+} \csc(\arctan x) & = \lim_{t \to 0^+} \csc t = \infty\\ \lim_{x \to \infty} \csc(\arctan x) & = \lim_{t \to \frac{\pi}{2}^-} \csc t = 1 \end{align*} $\csc(\arctan x)$ assumes every value in the interval $(1, \infty)$ by the Intermediate Value Theorem.

Since $\csc(\arctan x)$ is continuous on $(-\infty, 0)$, \begin{align*} \lim_{x \to 0^-} \csc(\arctan x) & = \lim_{t \to 0^-} \csc t = -\infty\\ \lim_{x \to -\infty} \csc(\arctan x) & = \lim_{t \to -\frac{\pi}{2}^+} \csc t = -1 \end{align*} $\csc(\arctan x)$ assumes every value in the interval $(-\infty, -1)$ by the Intermediate Value Theorem.

Therefore, the range of $\csc(\arctan x)$ is $(-\infty, - 1) \cup (1, \infty)$.

Answer 2

Domain is the possible set of values of $x$ for which f(x) is real and finite

For $x<0$

$-\dfrac\pi2<\tan^{-1}x<0\implies-\infty< f(x)<-1$

Similarly for $x>0$