The problem asks to express $\csc(t)$ in terms of $\cot(t)$ given that the terminal point of $t$ is in Quadrant III. Let's work through this step by step.

Given:

• The expression to solve is $\csc(t)$ in terms of $\cot(t)$.
• The angle $t$ lies in Quadrant III.
• We know that $\csc(t) = \frac{1}{\sin(t)}$ and $\cot(t) = \frac{\cos(t)}{\sin(t)}$.

Key Trigonometric Relationships:

In general: $\sin^2(t) + \cos^2(t) = 1$ In terms of $\cot(t)$: $\cot^2(t) = \frac{\cos^2(t)}{\sin^2(t)} \Rightarrow \sin^2(t) = \frac{1}{1 + \cot^2(t)}$

Step-by-Step Solution:

1. Find $\sin(t)$ in terms of $\cot(t)$: $\sin^2(t) = \frac{1}{1 + \cot^2(t)}$ $\sin(t) = \pm \frac{1}{\sqrt{1 + \cot^2(t)}}$

   Since we are in Quadrant III, where sine is negative, we choose the negative root: $\sin(t) = -\frac{1}{\sqrt{1 + \cot^2(t)}}$

2. Express $\csc(t)$: By definition, $\csc(t) = \frac{1}{\sin(t)}$. Substituting the expression for $\sin(t)$: $\csc(t) = -\sqrt{1 + \cot^2(t)}$

Thus, the answer is: $\csc(t) = -\sqrt{1 + \cot^2(t)}$

Would you like more details or have any questions about this solution?

Here are some related questions for further exploration:

1. How does the sign of $\csc(t)$ change in each quadrant?
2. What is the relationship between $\sec(t)$ and $\tan(t)$?
3. Can you derive $\cos(t)$ in terms of $\cot(t)$ in Quadrant III?
4. How does the Pythagorean identity apply to other trigonometric functions?
5. What is the significance of the sign of $\cot(t)$ in different quadrants?

Tip: Always consider the quadrant when determining the sign of trigonometric functions!