`(1-sinA+cosA)^(2)` is equal to ?

Options

A. `2(1-cosA)(1+sinA)`
B. `2(1-sinA)(1+cosA)`
C. `2(1-cosA)(1-sinA)`
D. None of the above

Explanation

To solve the expression \((1 - \sin A + \cos A)^2\), we will expand it using the identity for the square of a trinomial.

### Step-by-Step Solution:

1. **Identify the expression**:
We start with the expression:
\[ x = (1 - \sin A + \cos A)^2 \]

2. **Use the trinomial expansion**:
The formula for the square of a trinomial \((a + b + c)^2\) is:
\[ a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \]
Here, we can identify:
- \(a = 1\)
- \(b = -\sin A\)
- \(c = \cos A\)

3. **Apply the expansion**:
Now we substitute \(a\), \(b\), and \(c\) into the expansion:
\[ x = 1^2 + (-\sin A)^2 + (\cos A)^2 + 2(1)(-\sin A) + 2(-\sin A)(\cos A) + 2(\cos A)(1) \]

4. **Calculate each term**:
- \(1^2 = 1\)
- \((- \sin A)^2 = \sin^2 A\)
- \((\cos A)^2 = \cos^2 A\)
- \(2(1)(-\sin A) = -2\sin A\)
- \(2(-\sin A)(\cos A) = -2\sin A \cos A\)
- \(2(\cos A)(1) = 2\cos A\)

5. **Combine the terms**:
Now we combine all the terms:
\[ x = 1 + \sin^2 A + \cos^2 A - 2\sin A - 2\sin A \cos A + 2\cos A \]

6. **Use the Pythagorean identity**:
We know from the Pythagorean identity that:
\[ \sin^2 A + \cos^2 A = 1 \]
Therefore, we can substitute this into our expression:
\[ x = 1 + 1 - 2\sin A - 2\sin A \cos A + 2\cos A \]
Simplifying this gives:
\[ x = 2 - 2\sin A - 2\sin A \cos A + 2\cos A \]

7. **Factor out the common terms**:
We can factor out a 2 from the expression:
\[ x = 2(1 - \sin A - \sin A \cos A + \cos A) \]

8. **Rearranging the expression**:
Rearranging gives us:
\[ x = 2(1 - \sin A)(1 + \cos A) \]

### Final Result:
Thus, the expression \((1 - \sin A + \cos A)^2\) simplifies to:
\[ \boxed{2(1 - \sin A)(1 + \cos A)} \]