`cos^(2)(A-B)+cos^(2)B-2cos(A-B).cosA.cosB=?`

Options

A. `cos^(2)A`
B. `sin^(2)A`
C. `tan^(2)A`
D. `cot^(2)A`

Answer

To solve the equation \( \cos^2(A-B) + \cos^2B - 2\cos(A-B)\cos A \cos B \), we can follow these steps:

### Step 1: Substitute \( B = 0 \)
To simplify the expression, we can set \( B = 0 \). This gives us:
\[ \cos^2(A-0) + \cos^2(0) - 2\cos(A-0)\cos A \cos(0) \]
This simplifies to:
\[ \cos^2 A + \cos^2 0 - 2\cos A \cos A \cdot 1 \]
Since \( \cos 0 = 1 \), we have:
\[ \cos^2 A + 1 - 2\cos^2 A \]

### Step 2: Combine Like Terms
Now, we can combine the terms:
\[ 1 + \cos^2 A - 2\cos^2 A = 1 - \cos^2 A \]

### Step 3: Use the Pythagorean Identity
We know from the Pythagorean identity that:
\[ \sin^2 A = 1 - \cos^2 A \]
Thus, we can replace \( 1 - \cos^2 A \) with \( \sin^2 A \):
\[ 1 - \cos^2 A = \sin^2 A \]

### Final Result
Therefore, the expression simplifies to:
\[ \sin^2 A \]

### Conclusion
The answer to the question \( \cos^2(A-B) + \cos^2B - 2\cos(A-B)\cos A \cos B \) is \( \sin^2 A \).

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