1. **State the problem:** Prove that $$\frac{\tan x + \sec x - 1}{\tan x - \sec x + 1} = \frac{1 + \sin x}{\cos x}.$$\n\n2. **Rewrite tangent and secant in terms of sine and cosine:**\n$$\tan x = \frac{\sin x}{\cos x}, \quad \sec x = \frac{1}{\cos x}.$$\n\n3. **Substitute into the left-hand side (LHS):**\n$$\frac{\frac{\sin x}{\cos x} + \frac{1}{\cos x} - 1}{\frac{\sin x}{\cos x} - \frac{1}{\cos x} + 1}.$$\n\n4. **Combine terms over common denominator $\cos x$ in numerator and denominator:**\nNumerator: $$\frac{\sin x + 1 - \cos x}{\cos x}.$$\nDenominator: $$\frac{\sin x - 1 + \cos x}{\cos x}.$$\n\n5. **Rewrite LHS as:**\n$$\frac{\frac{\sin x + 1 - \cos x}{\cos x}}{\frac{\sin x - 1 + \cos x}{\cos x}} = \frac{\sin x + 1 - \cos x}{\sin x - 1 + \cos x}.$$\n\n6. **Simplify the denominator by factoring out a negative sign:**\n$$\sin x - 1 + \cos x = - (1 - \sin x - \cos x).$$\n\n7. **Rewrite LHS as:**\n$$\frac{\sin x + 1 - \cos x}{- (1 - \sin x - \cos x)} = - \frac{\sin x + 1 - \cos x}{1 - \sin x - \cos x}.$$\n\n8. **Note that:**\n$$1 - \sin x - \cos x = (1 - \sin x) - \cos x,$$\nand\n$$\sin x + 1 - \cos x = (1 + \sin x) - \cos x.$$\n\n9. **Rewrite numerator and denominator to see if they relate:**\nNumerator: $$1 + \sin x - \cos x,$$\nDenominator: $$1 - \sin x - \cos x.$$\n\n10. **Multiply numerator and denominator by the conjugate of the denominator's denominator to simplify:**\nMultiply numerator and denominator by $$1 + \sin x + \cos x.$$\n\n11. **Calculate numerator:**\n$$(1 + \sin x - \cos x)(1 + \sin x + \cos x) = (1 + \sin x)^2 - \cos^2 x = 1 + 2\sin x + \sin^2 x - \cos^2 x.$$\n\n12. **Calculate denominator:**\n$$(1 - \sin x - \cos x)(1 + \sin x + \cos x) = 1 - (\sin x + \cos x)^2 = 1 - (\sin^2 x + 2\sin x \cos x + \cos^2 x) = 1 - 1 - 2\sin x \cos x = -2\sin x \cos x.$$\n\n13. **Substitute back:**\n$$- \frac{1 + 2\sin x + \sin^2 x - \cos^2 x}{-2\sin x \cos x} = \frac{1 + 2\sin x + \sin^2 x - \cos^2 x}{2\sin x \cos x}.$$\n\n14. **Use identity $\sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x) = -\cos 2x$ and $1 - \cos 2x = 2\sin^2 x$ to simplify numerator:**\n$$1 + 2\sin x + \sin^2 x - \cos^2 x = (1 - \cos^2 x + \sin^2 x) + 2\sin x = 2\sin^2 x + 2\sin x.$$\n\n15. **Factor numerator:**\n$$2\sin x (\sin x + 1).$$\n\n16. **Rewrite entire expression:**\n$$\frac{2\sin x (\sin x + 1)}{2\sin x \cos x} = \frac{\sin x + 1}{\cos x}.$$\n\n17. **This matches the right-hand side (RHS):**\n$$\frac{1 + \sin x}{\cos x}.$$\n\n**Final answer:** The identity is proven: $$\frac{\tan x + \sec x - 1}{\tan x - \sec x + 1} = \frac{1 + \sin x}{\cos x}.$$