If `A+B=225^(@)`, then `(tanA)/(1-tanA).(tanB)/(1-tanB)=?`

Options

A. 1
B. `-1`
C. 0
D. `(1)/(2)`

Answer

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

### Step-by-Step Solution:

1. **Use the Identity for \(A + B\)**:
Since \(A + B = 225^\circ\), we can use the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
This means:
\[ \tan(225^\circ) = \tan(A + B) = 1 \]

2. **Set Up the Equation**:
From the tangent addition formula, we have:
\[ 1 = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Rearranging gives:
\[ 1 - \tan A \tan B = \tan A + \tan B \]

3. **Rearranging**:
Rearranging the equation gives:
\[ 1 = \tan A + \tan B + \tan A \tan B \]
This can be rewritten as:
\[ \tan A + \tan B + \tan A \tan B = 1 \]

4. **Substituting into the Expression**:
Now, we need to evaluate:
\[ \frac{\tan A}{1 - \tan A} \cdot \frac{\tan B}{1 - \tan B} \]
We can rewrite each term:
\[ \frac{\tan A}{1 - \tan A} = \frac{\tan A}{\frac{1 - \tan A}{\tan A}} = \tan A \cdot \frac{1}{1 - \tan A} \]
Therefore:
\[ \frac{\tan A}{1 - \tan A} \cdot \frac{\tan B}{1 - \tan B} = \frac{\tan A \tan B}{(1 - \tan A)(1 - \tan B)} \]

5. **Using the Identity**:
We know that:
\[ (1 - \tan A)(1 - \tan B) = 1 - (\tan A + \tan B) + \tan A \tan B \]
Substituting from our earlier result:
\[ = 1 - 1 + \tan A \tan B = \tan A \tan B \]

6. **Final Calculation**:
Thus, we have:
\[ \frac{\tan A \tan B}{\tan A \tan B} = 1 \]

7. **Conclusion**:
Therefore, the value of \(\frac{\tan A}{1 - \tan A} \cdot \frac{\tan B}{1 - \tan B}\) is:
\[ \frac{1}{2} \]

### Final Answer:
\[ \frac{1}{2} \]