If A+B`=225^(@)` then the value of
`(cotA)/(1+cotA).(cotB)/(1+cotB)` is

Options

A. 2
B. `1//2`
C. 3
D. `1//3`

Answer

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 1: Rewrite cotangent in terms of tangent
We know that

\[ \cot A = \frac{1}{\tan A} \quad \text{and} \quad \cot B = \frac{1}{\tan B} \]

Substituting these into the expression, we have:

\[ \frac{\cot A}{1 + \cot A} = \frac{\frac{1}{\tan A}}{1 + \frac{1}{\tan A}} = \frac{1}{\tan A + 1} \]

Similarly,

\[ \frac{\cot B}{1 + \cot B} = \frac{\frac{1}{\tan B}}{1 + \frac{1}{\tan B}} = \frac{1}{\tan B + 1} \]

### Step 2: Combine the two fractions
Now, we can combine the two fractions:

\[ \frac{\cot A}{1 + \cot A} \cdot \frac{\cot B}{1 + \cot B} = \frac{1}{\tan A + 1} \cdot \frac{1}{\tan B + 1} = \frac{1}{(\tan A + 1)(\tan B + 1)} \]

### Step 3: Use the angle sum identity
Since \( A + B = 225^\circ \), we can use the tangent addition formula:

\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

Since \( A + B = 225^\circ \), we have:

\[ \tan 225^\circ = 1 \]

Thus,

\[ 1 = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

### Step 4: Rearranging the equation
Cross-multiplying gives:

\[ 1 - \tan A \tan B = \tan A + \tan B \]

Rearranging this, we get:

\[ \tan A + \tan B + \tan A \tan B = 1 \]

### Step 5: Substitute back into the expression
Now, substituting this back into our expression:

\[ (\tan A + 1)(\tan B + 1) = \tan A \tan B + \tan A + \tan B + 1 \]

From the previous step, we know that \( \tan A + \tan B + \tan A \tan B = 1 \). Thus:

\[ (\tan A + 1)(\tan B + 1) = 1 + 1 = 2 \]

### Step 6: Final value
Now substituting this back into our expression gives:

\[ \frac{1}{(\tan A + 1)(\tan B + 1)} = \frac{1}{2} \]

Thus, the value of

\[ \frac{\cot A}{1 + \cot A} \cdot \frac{\cot B}{1 + \cot B} = \frac{1}{2} \]

### Final Answer
The final answer is:

\[ \frac{1}{2} \]