1. The problem asks to find the value of $$\frac{\sin \alpha + \cos \alpha}{\sin \alpha - \cos \alpha}$$ given that $$\tan \alpha = \frac{4}{5}$$. 2. Recall the identity $$\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$$. Since $$\tan \alpha = \frac{4}{5}$$, we can write: $$\frac{\sin \alpha}{\cos \alpha} = \frac{4}{5}$$ which implies $$\sin \alpha = \frac{4}{5} \cos \alpha$$. 3. Substitute $$\sin \alpha = \frac{4}{5} \cos \alpha$$ into the expression: $$\frac{\sin \alpha + \cos \alpha}{\sin \alpha - \cos \alpha} = \frac{\frac{4}{5} \cos \alpha + \cos \alpha}{\frac{4}{5} \cos \alpha - \cos \alpha}$$. 4. Factor $$\cos \alpha$$ out of numerator and denominator: $$= \frac{\cos \alpha \left( \frac{4}{5} + 1 \right)}{\cos \alpha \left( \frac{4}{5} - 1 \right)}$$. 5. Cancel $$\cos \alpha$$ (assuming $$\cos \alpha \neq 0$$): $$= \frac{\cancel{\cos \alpha} \left( \frac{4}{5} + 1 \right)}{\cancel{\cos \alpha} \left( \frac{4}{5} - 1 \right)}$$. 6. Simplify inside the parentheses: $$\frac{4}{5} + 1 = \frac{4}{5} + \frac{5}{5} = \frac{9}{5}$$ $$\frac{4}{5} - 1 = \frac{4}{5} - \frac{5}{5} = -\frac{1}{5}$$. 7. Substitute back: $$= \frac{\frac{9}{5}}{-\frac{1}{5}} = \frac{9}{5} \times \left(-5\right) = -9$$. 8. Therefore, the value of the expression is $$-9$$. Answer: C) -9