1. **State the problem:** Verify the identity $$(1 - \sin A + \cos A)^2 = 2(1 - \sin A)(1 + \cos A)$$. 2. **Expand the left side:** Use the formula for squaring a trinomial $$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$. Let $a = 1$, $b = -\sin A$, and $c = \cos A$. $$\begin{aligned} (1 - \sin A + \cos A)^2 &= 1^2 + (-\sin A)^2 + (\cos A)^2 + 2(1)(-\sin A) + 2(-\sin A)(\cos A) + 2(1)(\cos A) \\ &= 1 + \sin^2 A + \cos^2 A - 2\sin A - 2\sin A \cos A + 2\cos A \end{aligned}$$ 3. **Simplify using the Pythagorean identity:** $\sin^2 A + \cos^2 A = 1$. $$1 + 1 - 2\sin A - 2\sin A \cos A + 2\cos A = 2 - 2\sin A - 2\sin A \cos A + 2\cos A$$ 4. **Expand the right side:** $$2(1 - \sin A)(1 + \cos A) = 2(1 + \cos A - \sin A - \sin A \cos A) = 2 + 2\cos A - 2\sin A - 2\sin A \cos A$$ 5. **Compare both sides:** Left side: $$2 - 2\sin A - 2\sin A \cos A + 2\cos A$$ Right side: $$2 + 2\cos A - 2\sin A - 2\sin A \cos A$$ They are identical. **Final answer:** The identity is true.