To find the exact value of $\csc A$ given $\cot A = -\frac{5}{4}$ with angle $A$ in Quadrant IV, follow these steps:

1. Set up the right triangle based on $\cot A$:

   • Since $\cot A = -\frac{5}{4}$, we know that $\cot A = \frac{\text{adjacent}}{\text{opposite}}$.
   • In this case, the adjacent side is $5$ and the opposite side is $-4$ (negative because we are in Quadrant IV, where sine values are negative).

2. Use the Pythagorean theorem to find the hypotenuse: $\text{hypotenuse} = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2}$ Substituting in the values: $\text{hypotenuse} = \sqrt{(-4)^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}$

3. Determine $\csc A$:

   • $\csc A = \frac{\text{hypotenuse}}{\text{opposite}}$.
   • Substituting the values, we get: $\csc A = \frac{\sqrt{41}}{-4} = -\frac{\sqrt{41}}{4}$

So, the exact value of $\csc A$ in simplest radical form with a rational denominator is: $\csc A = -\frac{\sqrt{41}}{4}$

Would you like a breakdown of any part of this solution?

Here are 5 related questions:

1. How would you find $\sec A$ given $\cot A = -\frac{5}{4}$ and $A$ in Quadrant IV?
2. What is the value of $\sin A$ for the same conditions?
3. How does knowing the quadrant affect the sign of trigonometric functions?
4. How can you verify that this solution is correct using another trigonometric identity?
5. How would the answer change if $A$ were in Quadrant II?

Tip: In trigonometry, always pay close attention to the quadrant, as it determines the sign of the function values.