1. **State the problem:** Find the inverse Laplace transform of $$\frac{a \cos(b) - s \sin(b)}{s^2 + a^2}$$. 2. **Recall the Laplace transform formulas:** - The Laplace transform of $$\cos(at)$$ is $$\frac{s}{s^2 + a^2}$$. - The Laplace transform of $$\sin(at)$$ is $$\frac{a}{s^2 + a^2}$$. 3. **Rewrite the given expression:** $$\frac{a \cos(b) - s \sin(b)}{s^2 + a^2} = \cos(b) \cdot \frac{a}{s^2 + a^2} - \sin(b) \cdot \frac{s}{s^2 + a^2}$$ 4. **Identify the inverse transforms:** - $$\mathcal{L}^{-1}\left\{\frac{a}{s^2 + a^2}\right\} = \sin(at)$$ - $$\mathcal{L}^{-1}\left\{\frac{s}{s^2 + a^2}\right\} = \cos(at)$$ 5. **Apply linearity of inverse Laplace transform:** $$\mathcal{L}^{-1}\left\{\frac{a \cos(b) - s \sin(b)}{s^2 + a^2}\right\} = \cos(b) \sin(at) - \sin(b) \cos(at)$$ 6. **Use the sine difference identity:** $$\sin(at - b) = \sin(at) \cos(b) - \cos(at) \sin(b)$$ 7. **Final answer:** $$\boxed{\sin(at - b)}$$