Prove that `cos^(4)A-sin^(4)A=cos^(2)A-sin^(2)A`.

Answer

To prove that \( \cos^4 A - \sin^4 A = \cos^2 A - \sin^2 A \), we will start with the left-hand side (LHS) and manipulate it to show that it equals the right-hand side (RHS).

### Step-by-Step Solution:

1. **Start with the LHS:**
\[ \text{LHS} = \cos^4 A - \sin^4 A \]

2. **Recognize the difference of squares:**
The expression \( \cos^4 A - \sin^4 A \) can be factored using the difference of squares formula \( a^2 - b^2 = (a-b)(a+b) \).
Here, let \( a = \cos^2 A \) and \( b = \sin^2 A \):
\[ \cos^4 A - \sin^4 A = (\cos^2 A - \sin^2 A)(\cos^2 A + \sin^2 A) \]

3. **Use the Pythagorean identity:**
We know from trigonometric identities that:
\[ \cos^2 A + \sin^2 A = 1 \]
Substituting this into our expression gives:
\[ \text{LHS} = (\cos^2 A - \sin^2 A)(1) \]

4. **Simplify the expression:**
Thus, we have:
\[ \text{LHS} = \cos^2 A - \sin^2 A \]

5. **Conclude the proof:**
Since we have shown that the left-hand side equals the right-hand side, we conclude:
\[ \cos^4 A - \sin^4 A = \cos^2 A - \sin^2 A \]

### Final Result:
\[ \text{LHS} = \text{RHS} \]
Therefore, the equation \( \cos^4 A - \sin^4 A = \cos^2 A - \sin^2 A \) is proven.