1. **State the problem:** Verify the trigonometric identity $1 - \sin A + \cos A = 2(1 - \sin A)(1 + \cos A)$.\n\n2. **Recall the formula and rules:** We will expand the right-hand side (RHS) and simplify both sides to check if they are equal. Use distributive property and Pythagorean identity $\sin^2 A + \cos^2 A = 1$.\n\n3. **Expand the RHS:** $$2(1 - \sin A)(1 + \cos A) = 2[(1)(1) + (1)(\cos A) - (\sin A)(1) - (\sin A)(\cos A)] = 2(1 + \cos A - \sin A - \sin A \cos A)$$\n\n4. **Distribute 2:** $$= 2 + 2\cos A - 2\sin A - 2\sin A \cos A$$\n\n5. **Compare with LHS:** LHS is $1 - \sin A + \cos A$.\n\n6. **Check if LHS equals RHS:** LHS = $1 - \sin A + \cos A$\nRHS = $2 + 2\cos A - 2\sin A - 2\sin A \cos A$\n\n7. **Subtract LHS from RHS:** $$2 + 2\cos A - 2\sin A - 2\sin A \cos A - (1 - \sin A + \cos A) = 2 + 2\cos A - 2\sin A - 2\sin A \cos A - 1 + \sin A - \cos A$$ $$= (2 - 1) + (2\cos A - \cos A) + (-2\sin A + \sin A) - 2\sin A \cos A = 1 + \cos A - \sin A - 2\sin A \cos A$$\n\n8. **Since the difference is not zero, the identity does not hold for all $A$.**\n\n**Final answer:** The given equation is not an identity; it is false in general.