Question

what is the range of $y = \csc(x)$?
\\(\circ\\) $y \leq -1$
\\(\circ\\) $y \geq 1$
\\(\circ\\) $y \leq -1$ or $y \geq 1$
\\(\circ\\) $-1 \leq y \leq 1$

Explanation:

Step1: Recall the definition of cosecant

The cosecant function is defined as \( \csc(x)=\frac{1}{\sin(x)} \).

Step2: Analyze the range of sine function

We know that the range of \( \sin(x) \) is \( - 1\leqslant\sin(x)\leqslant1 \), and \( \sin(x)
eq0 \) (because \( \csc(x) \) is undefined when \( \sin(x) = 0 \)).

Step3: Analyze the range of reciprocal

Let \( t=\sin(x) \), then \( y = \frac{1}{t} \), where \( - 1\leqslant t\leqslant1 \) and \( t
eq0 \).

• When \( 0\lt t\leqslant1 \), as \( t \) increases from \( 0 \) to \( 1 \), \( \frac{1}{t} \) decreases from \( +\infty \) to \( 1 \), so \( y=\frac{1}{t}\geqslant1 \).
• When \( - 1\leqslant t\lt0 \), as \( t \) increases from \( - 1 \) to \( 0 \), \( \frac{1}{t} \) increases from \( - \infty \) to \( - 1 \), so \( y=\frac{1}{t}\leqslant - 1 \).

Combining these two cases, the range of \( y = \csc(x) \) is \( y\leqslant - 1 \) or \( y\geqslant1 \).

Answer:

\( y\leqslant - 1 \) or \( y\geqslant1 \) (the third option)