I'm having trouble understanding how my textbook derived the range for the trig function $\csc{\theta}$.
It starts with an angle $\theta$ in standard position in quadrant I increasing from near $0^o$ toward $90^o$.
The hypotenuse, $r$, stays the same throughout. As the angle measure increases, $y$ increases as well but never exceeds $r$, so $y\leq r$. (I assume this is because the hypotenuse of a right triangle is always the largest side).
Dividing both sides of the inequality, $y\leq r$, by $y$ gives: $1\leq \frac{r}{y}$. (I'm guessing both sides are divided by $y$ because $\csc{\theta}$ is defined as $\frac{r}{y}$).
$1\leq \frac{r}{y}$ can be rewritten as $1 \leq \csc{\theta}$.
Therefore, the range of cosecant in quadrants I and II (when $y$ is positive) is: $1 \leq \csc{\theta}$.
I run into problems/confusion, however, when I try to find the range for $\csc{\theta}$ in quadrants III and IV.
Again, there is a figure of an angle $\theta$ in standard position but this time in quadrant IV.
$r$ is the largest side since it's the hypotenuse, but because quadrant IV is below the x-axis then $y$ must be negative, so $-y \leq r$.
However, this time when I divide both sides by $y$ I get:
$-1 \leq \frac{r}{y}$
$-1 \leq \csc{\theta}$
I know this is incorrect and the range of cosecant should be going to negative infinity: $\csc{\theta} \leq -1$.
I'm not sure what I'm doing wrong.
Since $y$ is negative, when you divide by it, you have to reverse the inequality.