If `cot^2Acot^2B=3,` then the value of `(2-cos2A)(2-cos2B)` is ____

Explanation

To solve the problem, we start with the given equation:

**Given:**
\[ \cot^2 A \cot^2 B = 3 \]

We need to find the value of:
\[ (2 - \cos 2A)(2 - \cos 2B) \]

### Step 1: Express \(\cos 2A\) and \(\cos 2B\)

Using the double angle formula for cosine, we can express \(\cos 2A\) and \(\cos 2B\) in terms of sine:
\[ \cos 2A = 1 - 2\sin^2 A \]
\[ \cos 2B = 1 - 2\sin^2 B \]

### Step 2: Substitute into the expression

Now substitute these into the expression we need to evaluate:
\[ (2 - \cos 2A)(2 - \cos 2B) = (2 - (1 - 2\sin^2 A))(2 - (1 - 2\sin^2 B)) \]

This simplifies to:
\[ (2 - 1 + 2\sin^2 A)(2 - 1 + 2\sin^2 B) = (1 + 2\sin^2 A)(1 + 2\sin^2 B) \]

### Step 3: Expand the expression

Now we can expand this product:
\[ (1 + 2\sin^2 A)(1 + 2\sin^2 B) = 1 + 2\sin^2 A + 2\sin^2 B + 4\sin^2 A \sin^2 B \]

### Step 4: Relate \(\sin^2 A\) and \(\sin^2 B\) to \(\cot^2 A\) and \(\cot^2 B\)

From the given equation \(\cot^2 A \cot^2 B = 3\), we can express \(\cot^2 A\) and \(\cot^2 B\) in terms of sine and cosine:
\[ \cot^2 A = \frac{\cos^2 A}{\sin^2 A}, \quad \cot^2 B = \frac{\cos^2 B}{\sin^2 B} \]
Thus,
\[ \cot^2 A \cot^2 B = \frac{\cos^2 A \cos^2 B}{\sin^2 A \sin^2 B} = 3 \]
This implies:
\[ \cos^2 A \cos^2 B = 3 \sin^2 A \sin^2 B \]

### Step 5: Substitute and simplify

Using the identity \(\cos^2 A = 1 - \sin^2 A\) and \(\cos^2 B = 1 - \sin^2 B\), we substitute:
\[ (1 - \sin^2 A)(1 - \sin^2 B) = 3 \sin^2 A \sin^2 B \]
Expanding gives:
\[ 1 - \sin^2 A - \sin^2 B + \sin^2 A \sin^2 B = 3 \sin^2 A \sin^2 B \]
Rearranging leads to:
\[ 1 - \sin^2 A - \sin^2 B - 2\sin^2 A \sin^2 B = 0 \]
This can be rewritten as:
\[ \sin^2 A + \sin^2 B + 2\sin^2 A \sin^2 B = 1 \]

### Step 6: Multiply by 2

Now, multiply the entire equation by 2:
\[ 2\sin^2 A + 2\sin^2 B + 4\sin^2 A \sin^2 B = 2 \]

### Step 7: Substitute back into the expression

Now, we can substitute this result back into our expanded expression:
\[ 1 + 2\sin^2 A + 2\sin^2 B + 4\sin^2 A \sin^2 B = 1 + 2 = 3 \]

### Final Answer

Thus, the value of \((2 - \cos 2A)(2 - \cos 2B)\) is:
\[ \boxed{3} \]