1. **State the problem:** We need to find the value of $k$ in the equation $$\frac{\cot A}{1 + \csc A} - \frac{\cot A}{1 - \csc A} = \frac{k}{\cos A}.$$\n\n2. **Recall definitions and identities:** \n- $\cot A = \frac{\cos A}{\sin A}$\n- $\csc A = \frac{1}{\sin A}$\n- We will simplify the left side and express it in terms of $\cos A$ and $\sin A$.\n\n3. **Rewrite the left side:**\n$$\frac{\cot A}{1 + \csc A} - \frac{\cot A}{1 - \csc A} = \cot A \left( \frac{1}{1 + \csc A} - \frac{1}{1 - \csc A} \right).$$\n\n4. **Find common denominator inside parentheses:**\n$$\frac{1}{1 + \csc A} - \frac{1}{1 - \csc A} = \frac{(1 - \csc A) - (1 + \csc A)}{(1 + \csc A)(1 - \csc A)} = \frac{1 - \csc A - 1 - \csc A}{1 - (\csc A)^2} = \frac{-2 \csc A}{1 - \csc^2 A}.$$\n\n5. **Use Pythagorean identity:**\nSince $\csc^2 A = 1 + \cot^2 A$, then\n$$1 - \csc^2 A = 1 - (1 + \cot^2 A) = -\cot^2 A.$$\n\n6. **Substitute back:**\n$$\frac{-2 \csc A}{1 - \csc^2 A} = \frac{-2 \csc A}{-\cot^2 A} = \frac{2 \csc A}{\cot^2 A}.$$\n\n7. **Now the left side becomes:**\n$$\cot A \times \frac{2 \csc A}{\cot^2 A} = 2 \csc A \times \frac{\cot A}{\cot^2 A} = 2 \csc A \times \frac{1}{\cot A} = 2 \csc A \tan A.$$\n\n8. **Express $\csc A$ and $\tan A$ in terms of sine and cosine:**\n$$2 \csc A \tan A = 2 \times \frac{1}{\sin A} \times \frac{\sin A}{\cos A} = \frac{2}{\cos A}.$$\n\n9. **Therefore, the left side simplifies to:**\n$$\frac{2}{\cos A}.$$\n\n10. **Compare with the right side:**\n$$\frac{k}{\cos A} = \frac{2}{\cos A} \implies k = 2.$$\n\n**Final answer:** $k = 2$