1. **State the problem:** We need to find the values of $A$ and $B$ such that $$\csc\left(\arctan\left(-\frac{5}{12}\right)\right) = \frac{A}{B}.$$\n\n2. **Recall definitions:** \n- $\arctan\left(-\frac{5}{12}\right)$ is the angle $\theta$ whose tangent is $-\frac{5}{12}$.\n- $\csc(\theta) = \frac{1}{\sin(\theta)}$.\n\n3. **Find $\sin(\theta)$ given $\tan(\theta) = -\frac{5}{12}$:**\n- Use the identity $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$.\n- Let $\sin(\theta) = y$ and $\cos(\theta) = x$. Then $\frac{y}{x} = -\frac{5}{12}$.\n- We can represent this as a right triangle with opposite side $-5$ and adjacent side $12$.\n\n4. **Calculate hypotenuse:**\n$$\text{hypotenuse} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13.$$\n\n5. **Find $\sin(\theta)$ and $\cos(\theta)$:**\n$$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{-5}{13}, \quad \cos(\theta) = \frac{12}{13}.$$\n\n6. **Calculate $\csc(\theta)$:**\n$$\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{1}{-\frac{5}{13}} = -\frac{13}{5}.$$\n\n7. **Identify $A$ and $B$:**\n$$\frac{A}{B} = -\frac{13}{5} \implies A = -13, \quad B = 5.$$