If `A+B=225^(@)`, then find the value of `(cotA)/(1+cotA)xx(cot B)/(1+cotB)`

Explanation

To solve the problem, we need to find the value of \(\frac{\cot A}{1 + \cot A} \cdot \frac{\cot B}{1 + \cot B}\) given that \(A + B = 225^\circ\).

### Step-by-Step Solution:

1. **Use the Given Information**:
We know that \(A + B = 225^\circ\).

2. **Apply the Tangent Addition Formula**:
From the angle sum identity for tangent, we have:
\[ \tan(A + B) = \tan(225^\circ) \]
We know that \(\tan(225^\circ) = \tan(180^\circ + 45^\circ) = \tan(45^\circ) = 1\). Therefore:
\[ \tan(A + B) = 1 \]

3. **Express in Terms of Tangents**:
Using the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Setting this equal to 1 gives us:
\[ \frac{\tan A + \tan B}{1 - \tan A \tan B} = 1 \]

4. **Cross Multiply**:
Cross multiplying gives:
\[ \tan A + \tan B = 1 - \tan A \tan B \]

5. **Rearrange the Equation**:
Rearranging the equation leads to:
\[ \tan A + \tan B + \tan A \tan B = 1 \]

6. **Substituting into the Original Expression**:
We need to find:
\[ \frac{\cot A}{1 + \cot A} \cdot \frac{\cot B}{1 + \cot B} \]
We know that \(\cot A = \frac{1}{\tan A}\) and \(\cot B = \frac{1}{\tan B}\). Thus:
\[ \frac{\cot A}{1 + \cot A} = \frac{\frac{1}{\tan A}}{1 + \frac{1}{\tan A}} = \frac{1}{\tan A + 1} \]
Similarly for \(\cot B\):
\[ \frac{\cot B}{1 + \cot B} = \frac{1}{\tan B + 1} \]

7. **Combine the Two Expressions**:
Therefore:
\[ \frac{\cot A}{1 + \cot A} \cdot \frac{\cot B}{1 + \cot B} = \frac{1}{(\tan A + 1)(\tan B + 1)} \]

8. **Expand the Denominator**:
Expanding the denominator:
\[ (\tan A + 1)(\tan B + 1) = \tan A \tan B + \tan A + \tan B + 1 \]
From our earlier result, we know that \(\tan A + \tan B + \tan A \tan B = 1\). Thus:
\[ \tan A \tan B + \tan A + \tan B + 1 = 1 + 1 = 2 \]

9. **Final Result**:
Therefore:
\[ \frac{\cot A}{1 + \cot A} \cdot \frac{\cot B}{1 + \cot B} = \frac{1}{2} \]

### Conclusion:
The value of \(\frac{\cot A}{1 + \cot A} \cdot \frac{\cot B}{1 + \cot B}\) is \(\frac{1}{2}\).