To solve this problem, let’s calculate the requested trigonometric values step by step:

Given:

1. $\csc(A) = \frac{-13}{5}$, and $A$ is in Quadrant IV.
2. $\sec(B) = -2$, and $B$ is in Quadrant II.

1. Calculating for $A$:

• $\sin(A)$: Since $\csc(A) = \frac{1}{\sin(A)}$, $\sin(A) = \frac{1}{\csc(A)} = \frac{1}{\frac{-13}{5}} = \frac{-5}{13}.$

• $\cos(A)$: Using the Pythagorean identity: $\sin^2(A) + \cos^2(A) = 1.$ Substituting $\sin(A) = \frac{-5}{13}$: $\left(\frac{-5}{13}\right)^2 + \cos^2(A) = 1 \implies \frac{25}{169} + \cos^2(A) = 1.$ $\cos^2(A) = 1 - \frac{25}{169} = \frac{144}{169}.$ Taking the square root, and since $A$ is in Quadrant IV ($\cos > 0$): $\cos(A) = \frac{12}{13}.$

• $\tan(A)$: Using $\tan(A) = \frac{\sin(A)}{\cos(A)}$: $\tan(A) = \frac{\frac{-5}{13}}{\frac{12}{13}} = \frac{-5}{12}.$

2. Calculating for $B$:

• $\cos(B)$: Since $\sec(B) = \frac{1}{\cos(B)}$, $\cos(B) = \frac{1}{\sec(B)} = \frac{1}{-2} = -\frac{1}{2}.$

• $\sin(B)$: Using the Pythagorean identity: $\sin^2(B) + \cos^2(B) = 1.$ Substituting $\cos(B) = -\frac{1}{2}$: $\sin^2(B) + \left(-\frac{1}{2}\right)^2 = 1 \implies \sin^2(B) + \frac{1}{4} = 1.$ $\sin^2(B) = 1 - \frac{1}{4} = \frac{3}{4}.$ Taking the square root, and since $B$ is in Quadrant II ($\sin > 0$): $\sin(B) = \frac{\sqrt{3}}{2}.$

• $\tan(B)$: Using $\tan(B) = \frac{\sin(B)}{\cos(B)}$: $\tan(B) = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -\sqrt{3}.$

Final Values:

1. $\sin(A) = \frac{-5}{13}$
2. $\cos(A) = \frac{12}{13}$
3. $\tan(A) = \frac{-5}{12}$
4. $\sin(B) = \frac{\sqrt{3}}{2}$
5. $\cos(B) = -\frac{1}{2}$
6. $\tan(B) = -\sqrt{3}$

Let me know if you need clarifications or details for any step! Here are five relative questions to expand:

1. How can trigonometric functions be determined for other quadrants?
2. What is the significance of signs in different quadrants for trigonometric values?
3. Can you explain how the Pythagorean identity works for any angle?
4. How are reciprocal trigonometric functions (like $\csc, \sec$) related to their primary counterparts?
5. How can these trigonometric concepts be applied in real-world problems?

Tip: Always double-check the quadrant to assign the correct