Let’s solve the given problem step by step.

Given Information:

1. For angle $A$:

   • $\csc(A) = -\frac{13}{5}$ ($A$ is in Quadrant IV).
   • Therefore, $\sin(A) = \frac{-5}{13}$ (reciprocal of $\csc(A)$).

2. For angle $B$:

   • $\sec(B) = -2$ ($B$ is in Quadrant II).
   • Therefore, $\cos(B) = -\frac{1}{2}$ (reciprocal of $\sec(B)$).

Step 1: Solve for $\cos(A)$ and $\tan(A)$:

Using the Pythagorean identity: $\sin^2(A) + \cos^2(A) = 1$ $\left(\frac{-5}{13}\right)^2 + \cos^2(A) = 1$ $\frac{25}{169} + \cos^2(A) = 1$ $\cos^2(A) = \frac{144}{169}$ $\cos(A) = \pm \frac{12}{13}$ Since $A$ is in Quadrant IV, $\cos(A) > 0$: $\cos(A) = \frac{12}{13}$

Now, $\tan(A)$ is: $\tan(A) = \frac{\sin(A)}{\cos(A)} = \frac{-\frac{5}{13}}{\frac{12}{13}} = -\frac{5}{12}$

Step 2: Solve for $\sin(B)$ and $\tan(B)$:

Using the Pythagorean identity: $\sin^2(B) + \cos^2(B) = 1$ $\sin^2(B) + \left(-\frac{1}{2}\right)^2 = 1$ $\sin^2(B) + \frac{1}{4} = 1$ $\sin^2(B) = \frac{3}{4}$ $\sin(B) = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$ Since $B$ is in Quadrant II, $\sin(B) > 0$: $\sin(B) = \frac{\sqrt{3}}{2}$

Now, $\tan(B)$ is: $\tan(B) = \frac{\sin(B)}{\cos(B)} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -\sqrt{3}$

Final Results:

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

Let me know if you'd like a detailed explanation for any part or help with other aspects of this problem!

Additional Questions to Explore:

1. How do the signs of trigonometric functions differ across quadrants?
2. What is the reciprocal identity relationship between $\csc$, $\sec$, and $\tan$?
3. How would the solution change if the given quadrants were different?
4. Can we verify these results using the unit circle?
5. How do these values translate to real-world applications of trigonometry?

Tip:

Always use the Pythagorean identities and the quadrant-specific signs to verify trigonometric values.