To solve the given equation [tex]\((1 - \sin A + \cos A)^2 = 2(1 - \sin A)(1 + \cos A)\)[/tex], we need to simplify both sides and then check if they are equivalent.
### Step 1: Simplify the Left-Hand Side (LHS)
Let's begin with the left-hand side:
[tex]\[
(1 - \sin A + \cos A)^2
\][/tex]
We will expand this by using the algebraic identity [tex]\((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = - \sin A\)[/tex], and [tex]\(c = \cos A\)[/tex]:
[tex]\[
(1 - \sin A + \cos A)^2 = 1^2 + (-\sin A)^2 + (\cos A)^2 + 2 \cdot 1 \cdot (-\sin A) + 2 \cdot (-\sin A) \cdot \cos A + 2 \cdot 1 \cdot \cos A
\][/tex]
Simplify each term:
[tex]\[
1 + \sin^2 A + \cos^2 A - 2 \sin A - 2 \sin A \cos A + 2 \cos A
\][/tex]
Recall that [tex]\(\sin^2 A + \cos^2 A = 1\)[/tex]:
[tex]\[
1 + 1 - 2 \sin A - 2 \sin A \cos A + 2 \cos A
\][/tex]
Combine like terms:
[tex]\[
2 - 2 \sin A - 2 \sin A \cos A + 2 \cos A
\][/tex]
Factor out the common factor of 2:
[tex]\[
2(1 - \sin A + \cos A - \sin A \cos A)
\][/tex]
### Step 2: Simplify the Right-Hand Side (RHS)
Now, let's simplify the right-hand side:
[tex]\[
2(1 - \sin A)(1 + \cos A)
\][/tex]
Apply the distributive property:
[tex]\[
2 [(1 \cdot 1) + (1 \cdot \cos A) + (-\sin A \cdot 1) + (-\sin A \cdot \cos A)]
\][/tex]
Simplify inside the bracket:
[tex]\[
2 [1 + \cos A - \sin A - \sin A \cos A]
\][/tex]
### Step 3: Compare LHS and RHS
We have simplified both sides:
[tex]\[
\text{LHS: } 2(1 - \sin A + \cos A - \sin A \cos A)
\][/tex]
[tex]\[
\text{RHS: } 2(1 + \cos A - \sin A - \sin A \cos A)
\][/tex]
Both expressions inside the brackets are the same:
[tex]\[
1 + \cos A - \sin A - \sin A \cos A
\][/tex]
Thus, the two sides are indeed equal. Therefore, the original equation holds true:
[tex]\[
(1 - \sin A + \cos A)^2 = 2(1 - \sin A)(1 + \cos A)
\][/tex]
This completes the verification of the given trigonometric identity.
