1. **State the problem:** Given $\cos A = \frac{3}{5}$ with $A$ acute, and $\sin B = \frac{5}{13}$ with $B$ obtuse, find $\cos(A-B) - \sin(A-B)$ without calculators. 2. **Recall formulas:** - $\cos(A-B) = \cos A \cos B + \sin A \sin B$ - $\sin(A-B) = \sin A \cos B - \cos A \sin B$ We want to find: $$\cos(A-B) - \sin(A-B) = (\cos A \cos B + \sin A \sin B) - (\sin A \cos B - \cos A \sin B)$$ 3. **Simplify the expression:** $$= \cos A \cos B + \sin A \sin B - \sin A \cos B + \cos A \sin B$$ Group terms: $$= \cos A \cos B + \cos A \sin B + \sin A \sin B - \sin A \cos B$$ 4. **Find $\sin A$ and $\cos B$ using Pythagorean identity:** Since $A$ is acute and $\cos A = \frac{3}{5}$, $$\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$ Since $B$ is obtuse and $\sin B = \frac{5}{13}$, Recall that for obtuse angles, $\sin B > 0$ and $\cos B < 0$. $$\cos B = -\sqrt{1 - \sin^2 B} = -\sqrt{1 - \left(\frac{5}{13}\right)^2} = -\sqrt{1 - \frac{25}{169}} = -\sqrt{\frac{144}{169}} = -\frac{12}{13}$$ 5. **Substitute values into the expression:** $$\cos A \cos B = \frac{3}{5} \times \left(-\frac{12}{13}\right) = -\frac{36}{65}$$ $$\cos A \sin B = \frac{3}{5} \times \frac{5}{13} = \frac{15}{65}$$ $$\sin A \sin B = \frac{4}{5} \times \frac{5}{13} = \frac{20}{65}$$ $$\sin A \cos B = \frac{4}{5} \times \left(-\frac{12}{13}\right) = -\frac{48}{65}$$ 6. **Calculate the sum:** $$-\frac{36}{65} + \frac{15}{65} + \frac{20}{65} - \left(-\frac{48}{65}\right) = -\frac{36}{65} + \frac{15}{65} + \frac{20}{65} + \frac{48}{65}$$ Combine numerators: $$= \frac{-36 + 15 + 20 + 48}{65} = \frac{47}{65}$$ **Final answer:** $$\cos(A-B) - \sin(A-B) = \frac{47}{65}$$