1. **State the problem:** Given $\sin(A+B) = 0.75$ and $\sin(A-B) = 0.43$, find the value of $\cos A \sin B$ to the nearest hundredth. 2. **Recall the sine addition and subtraction formulas:** $$\sin(A+B) = \sin A \cos B + \cos A \sin B$$ $$\sin(A-B) = \sin A \cos B - \cos A \sin B$$ 3. **Add the two equations:** $$\sin(A+B) + \sin(A-B) = (\sin A \cos B + \cos A \sin B) + (\sin A \cos B - \cos A \sin B)$$ $$0.75 + 0.43 = 2 \sin A \cos B$$ $$1.18 = 2 \sin A \cos B$$ $$\sin A \cos B = \frac{1.18}{2} = 0.59$$ 4. **Subtract the two equations:** $$\sin(A+B) - \sin(A-B) = (\sin A \cos B + \cos A \sin B) - (\sin A \cos B - \cos A \sin B)$$ $$0.75 - 0.43 = 2 \cos A \sin B$$ $$0.32 = 2 \cos A \sin B$$ $$\cos A \sin B = \frac{0.32}{2} = 0.16$$ 5. **Final answer:** $$\boxed{0.16}$$ This is the value of $\cos A \sin B$ to the nearest hundredth.