To find [tex]\(\csc(\theta)\)[/tex] given [tex]\(\cot(\theta) = 4\)[/tex] and that the terminal side of [tex]\(\theta\)[/tex] lies in quadrant III, follow these steps:
1. Express [tex]\(\cot(\theta)\)[/tex]:
[tex]\(\cot(\theta) = \frac{\text{adjacent}}{\text{opposite}}\)[/tex].
Given [tex]\(\cot(\theta) = 4\)[/tex], we can write this ratio as:
[tex]\[
\cot(\theta) = \frac{4}{1}
\][/tex]
This means the adjacent side is 4 and the opposite side is 1.
2. Adjust for quadrant III:
In quadrant III, both the x (adjacent) and y (opposite) coordinates are negative:
[tex]\[
\text{adjacent} = -4, \quad \text{opposite} = -1
\][/tex]
3. Find the hypotenuse:
Use the Pythagorean theorem to find the hypotenuse [tex]\(r\)[/tex]:
[tex]\[
r = \sqrt{(\text{adjacent})^2 + (\text{opposite})^2} = \sqrt{(-4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}
\][/tex]
4. Calculate [tex]\(\csc(\theta)\)[/tex]:
[tex]\(\csc(\theta)\)[/tex] is the reciprocal of [tex]\(\sin(\theta)\)[/tex], and [tex]\(\sin(\theta)\)[/tex] is the ratio of the opposite side to the hypotenuse:
[tex]\[
\sin(\theta) = \frac{-1}{\sqrt{17}}
\][/tex]
Therefore,
[tex]\[
\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{\sqrt{17}}{-1} = -\sqrt{17}
\][/tex]
So the exact, fully simplified value of [tex]\(\csc(\theta)\)[/tex] is:
[tex]\[
\csc(\theta) = -\sqrt{17}
\][/tex]