The value of `(sinA + cos A)^(2) + ( cos A - sin A)^(2)` is

Options

A. `0`
B. `1`
C. `2`
D. `3`

Answer

To solve the expression \((\sin A + \cos A)^{2} + (\cos A - \sin A)^{2}\), we can follow these steps:

### Step 1: Expand the squares
Using the formula \((a + b)^{2} = a^{2} + b^{2} + 2ab\) and \((a - b)^{2} = a^{2} + b^{2} - 2ab\), we can expand both terms:

\[ (\sin A + \cos A)^{2} = \sin^{2} A + \cos^{2} A + 2\sin A \cos A \]

\[ (\cos A - \sin A)^{2} = \cos^{2} A + \sin^{2} A - 2\sin A \cos A \]

### Step 2: Combine the expanded terms
Now, we can combine the two expanded expressions:

\[ (\sin^{2} A + \cos^{2} A + 2\sin A \cos A) + (\cos^{2} A + \sin^{2} A - 2\sin A \cos A) \]

### Step 3: Simplify the expression
Combine like terms:

\[ \sin^{2} A + \cos^{2} A + \cos^{2} A + \sin^{2} A + 2\sin A \cos A - 2\sin A \cos A \]

This simplifies to:

\[ 2\sin^{2} A + 2\cos^{2} A \]

### Step 4: Use the Pythagorean identity
We know from the Pythagorean identity that \(\sin^{2} A + \cos^{2} A = 1\). Therefore:

\[ 2(\sin^{2} A + \cos^{2} A) = 2 \cdot 1 = 2 \]

### Final Answer
Thus, the value of \((\sin A + \cos A)^{2} + (\cos A - \sin A)^{2}\) is:

\[ \boxed{2} \]