If `A+B=45^(@)`, then `(cotA-1) (cotB-1)` is equal to

Options

A. 3
B. `(1)/(2)`
C. `-1`
D. 2

Explanation

To solve the problem, we need to find the value of \((\cot A - 1)(\cot B - 1)\) given that \(A + B = 45^\circ\).

### Step-by-Step Solution:

1. **Use the cotangent addition formula**:
Since \(A + B = 45^\circ\), we can use the cotangent addition formula:
\[ \cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B} \]
Given that \(\cot 45^\circ = 1\), we have:
\[ 1 = \frac{\cot A \cot B - 1}{\cot A + \cot B} \]

2. **Set up the equation**:
From the equation above, we can cross-multiply to eliminate the fraction:
\[ \cot A + \cot B = \cot A \cot B - 1 \]

3. **Rearranging the equation**:
Rearranging gives us:
\[ \cot A \cot B - \cot A - \cot B - 1 = 0 \]

4. **Factor the equation**:
We can factor this equation as follows:
\[ (\cot A - 1)(\cot B - 1) = 2 \]
This can be derived by recognizing that if we let \(x = \cot A - 1\) and \(y = \cot B - 1\), then:
\[ xy = \cot A \cot B - \cot A - \cot B + 1 \]
and we know from our previous step that this equals 2.

5. **Conclusion**:
Therefore, we find that:
\[ (\cot A - 1)(\cot B - 1) = 2 \]

### Final Answer:
\[ (\cot A - 1)(\cot B - 1) = 2 \]