1. **State the problem:** We need to find $\sin(\alpha - \beta)$ given $\cos \alpha = \frac{12}{13}$ with $\alpha$ in quadrant IV, and $\tan \beta = \frac{5}{12}$ with $\beta$ in quadrant I. 2. **Recall the formula:** $$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$$ 3. **Find $\sin \alpha$:** Since $\cos \alpha = \frac{12}{13}$ and $\alpha$ is in quadrant IV, where sine is negative, $$\sin \alpha = -\sqrt{1 - \cos^2 \alpha} = -\sqrt{1 - \left(\frac{12}{13}\right)^2} = -\sqrt{1 - \frac{144}{169}} = -\sqrt{\frac{25}{169}} = -\frac{5}{13}$$ 4. **Find $\sin \beta$ and $\cos \beta$:** Given $\tan \beta = \frac{5}{12}$ and $\beta$ in quadrant I (both sine and cosine positive), use the identity: $$\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{5}{12}$$ Let $\cos \beta = x$, then $\sin \beta = \frac{5}{12} x$. Using $\sin^2 \beta + \cos^2 \beta = 1$: $$\left(\frac{5}{12} x\right)^2 + x^2 = 1 \Rightarrow \frac{25}{144} x^2 + x^2 = 1 \Rightarrow \frac{169}{144} x^2 = 1 \Rightarrow x^2 = \frac{144}{169}$$ Since $\beta$ is in quadrant I, $\cos \beta = \frac{12}{13}$. Then, $$\sin \beta = \frac{5}{12} \times \frac{12}{13} = \frac{5}{13}$$ 5. **Substitute values into the formula:** $$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta = \left(-\frac{5}{13}\right) \times \frac{12}{13} - \frac{12}{13} \times \frac{5}{13} = -\frac{60}{169} - \frac{60}{169} = -\frac{120}{169}$$ 6. **Final answer:** $$\sin(\alpha - \beta) = -\frac{120}{169}$$