1. Stating the problem: Verify if $$\frac{2\csc^2 A - 2\csc A \cot A}{-2\cot^2 A + 2\csc A \cot A} = \sec A$$. 2. Factor the numerator and denominator: Numerator: $$2\csc^2 A - 2\csc A \cot A = 2\csc A (\csc A - \cot A)$$ Denominator: $$-2\cot^2 A + 2\csc A \cot A = 2\cot A (\csc A - \cot A)$$ 3. Substitute these back into the fraction: $$\frac{2\csc A (\csc A - \cot A)}{2\cot A (\csc A - \cot A)}$$ 4. Cancel the common factor \(2(\csc A - \cot A)\) (assuming \(\csc A \neq \cot A\)): $$\frac{\csc A}{\cot A}$$ 5. Rewrite \(\cot A\) as \(\frac{\cos A}{\sin A}\) and \(\csc A\) as \(\frac{1}{\sin A}\): $$\frac{\frac{1}{\sin A}}{\frac{\cos A}{\sin A}} = \frac{1}{\sin A} \times \frac{\sin A}{\cos A} = \frac{1}{\cos A} = \sec A$$ 6. Conclusion: The given expression simplifies to \(\sec A\), so the equality holds. Final answer: $$\frac{2\csc^2 A - 2\csc A \cot A}{-2\cot^2 A + 2\csc A \cot A} = \sec A$$