1. We are asked to show that $$\frac{\cot^2 \alpha}{\csc \alpha - 1} = \csc \alpha + 1$$. 2. Recall the identities: - $$\cot^2 \alpha = \csc^2 \alpha - 1$$ - $$\csc \alpha = \frac{1}{\sin \alpha}$$ 3. Substitute $$\cot^2 \alpha$$ in the left-hand side (LHS): $$\frac{\cot^2 \alpha}{\csc \alpha - 1} = \frac{\csc^2 \alpha - 1}{\csc \alpha - 1}$$ 4. Factor the numerator as a difference of squares: $$\csc^2 \alpha - 1 = (\csc \alpha - 1)(\csc \alpha + 1)$$ 5. Substitute back: $$\frac{(\csc \alpha - 1)(\csc \alpha + 1)}{\csc \alpha - 1}$$ 6. Cancel the common factor $$\csc \alpha - 1$$ (assuming $$\csc \alpha \neq 1$$): $$\csc \alpha + 1$$ 7. This matches the right-hand side (RHS), so the identity is proven. Final answer: $$\frac{\cot^2 \alpha}{\csc \alpha - 1} = \csc \alpha + 1$$