1. **State the problem:** Prove that $$\frac{\sin A + \cos A - 1}{\sin A - \cos A + 1} = \frac{\cos A}{1 + \sin A}$$. 2. **Start with the left-hand side (LHS):** $$\frac{\sin A + \cos A - 1}{\sin A - \cos A + 1}$$. 3. **Multiply numerator and denominator by the conjugate of the denominator's denominator to simplify:** Multiply numerator and denominator by $$\sin A - \cos A - 1$$. 4. **Calculate numerator:** $$ (\sin A + \cos A - 1)(\sin A - \cos A - 1) $$ Expand using distributive property: $$= \sin^2 A - \sin A \cos A - \sin A + \cos A \sin A - \cos^2 A - \cos A - \sin A + \cos A + 1 $$ Simplify terms: $$= \sin^2 A - \cos^2 A - 2 \sin A + 1 $$ 5. **Calculate denominator:** $$ (\sin A - \cos A + 1)(\sin A - \cos A - 1) $$ This is a difference of squares: $$= (\sin A - \cos A)^2 - 1^2 = (\sin A - \cos A)^2 - 1 $$ Expand square: $$= \sin^2 A - 2 \sin A \cos A + \cos^2 A - 1 $$ 6. **Rewrite numerator and denominator using Pythagorean identity $$\sin^2 A + \cos^2 A = 1$$:** Numerator: $$ 1 - 2 \cos^2 A - 2 \sin A + 1 = 2 - 2 \cos^2 A - 2 \sin A $$ Denominator: $$ 1 - 2 \sin A \cos A + 1 - 1 = 1 - 2 \sin A \cos A $$ 7. **Simplify numerator:** $$ 2 - 2 \cos^2 A - 2 \sin A = 2(1 - \cos^2 A - \sin A) $$ Using identity $$1 - \cos^2 A = \sin^2 A$$: $$= 2(\sin^2 A - \sin A) = 2 \sin A (\sin A - 1) $$ 8. **Simplify denominator:** $$ 1 - 2 \sin A \cos A $$ 9. **Rewrite LHS:** $$ \frac{2 \sin A (\sin A - 1)}{1 - 2 \sin A \cos A} $$ 10. **Now consider the right-hand side (RHS):** $$ \frac{\cos A}{1 + \sin A} $$ 11. **Goal:** Show LHS = RHS. 12. **Try to simplify or transform LHS to RHS or vice versa.** 13. **Alternatively, cross-multiply original equation:** $$ (\sin A + \cos A - 1)(1 + \sin A) = \cos A (\sin A - \cos A + 1) $$ 14. **Expand both sides:** LHS: $$ \sin A (1 + \sin A) + \cos A (1 + \sin A) - 1 (1 + \sin A) = \sin A + \sin^2 A + \cos A + \cos A \sin A - 1 - \sin A $$ Simplify: $$ \sin A - \sin A = 0 $$ So LHS: $$ \sin^2 A + \cos A + \cos A \sin A - 1 $$ RHS: $$ \cos A \sin A - \cos^2 A + \cos A $$ 15. **Set LHS = RHS:** $$ \sin^2 A + \cos A + \cos A \sin A - 1 = \cos A \sin A - \cos^2 A + \cos A $$ 16. **Subtract $$\cos A \sin A + \cos A$$ from both sides:** $$ \sin^2 A - 1 = - \cos^2 A $$ 17. **Rewrite:** $$ \sin^2 A - 1 = - \cos^2 A $$ Add $$\cos^2 A$$ to both sides: $$ \sin^2 A + \cos^2 A - 1 = 0 $$ 18. **Use Pythagorean identity:** $$ \sin^2 A + \cos^2 A = 1 $$ So: $$ 1 - 1 = 0 $$ Which is true. 19. **Conclusion:** The original equation holds true. **Final answer:** $$\boxed{\frac{\sin A + \cos A - 1}{\sin A - \cos A + 1} = \frac{\cos A}{1 + \sin A}}$$