Let’s solve this problem step-by-step based on the given details:

Problem Breakdown:

1. For $A$: We are told that:

   • $\csc(A) = -\frac{13}{5}$, which means $\sin(A) = -\frac{5}{13}$ (since $\csc$ is the reciprocal of $\sin$).
   • $A$ terminates in Quadrant IV, where $\cos(A) > 0$ and $\sin(A) < 0$.

   Using the Pythagorean Identity: $\sin^2(A) + \cos^2(A) = 1$ Substituting $\sin(A) = -\frac{5}{13}$: $\left(-\frac{5}{13}\right)^2 + \cos^2(A) = 1$ $\frac{25}{169} + \cos^2(A) = 1$ $\cos^2(A) = 1 - \frac{25}{169} = \frac{144}{169}$ $\cos(A) = \sqrt{\frac{144}{169}} = \frac{12}{13} \quad (\text{positive in Quadrant IV})$

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

   Summary for $A$:

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

2. For $B$: We are told that:

   • $\sec(B) = -2$, which means $\cos(B) = -\frac{1}{2}$ (since $\sec$ is the reciprocal of $\cos$).
   • $B$ terminates in Quadrant II, where $\sin(B) > 0$ and $\cos(B) < 0$.

   Using the Pythagorean Identity: $\sin^2(B) + \cos^2(B) = 1$ Substituting $\cos(B) = -\frac{1}{2}$: $\sin^2(B) + \left(-\frac{1}{2}\right)^2 = 1$ $\sin^2(B) + \frac{1}{4} = 1$ $\sin^2(B) = 1 - \frac{1}{4} = \frac{3}{4}$ $\sin(B) = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \quad (\text{positive in Quadrant II})$

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

   Summary for $B$:

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

Completed Table:

Would you like a detailed explanation of any specific part or more problems like this?

Relative Questions:

1. How do we calculate the other trigonometric functions if only one is given?
2. Why does the quadrant affect the sign of the trigonometric values?
3. What is the relationship between $\sec$ and $\cos$, or $\csc$ and $\sin$?
4. How does the Pythagorean Identity help in finding missing values?
5. How do we determine the quadrant of the angle using trigonometric signs?

Tip: Always consider the quadrant to correctly