If `alpha+beta=(pi)/(4)`then value of `(1=tanalpha).(1+tanbeta)`is

Options

A. 1
B. 2
C. (-2)
D. Not defined

Answer

To solve the problem, we need to find the value of \( (1 + \tan \alpha)(1 + \tan \beta) \) given that \( \alpha + \beta = \frac{\pi}{4} \).

### Step-by-step Solution:

1. **Use the given information:**
We know that \( \alpha + \beta = \frac{\pi}{4} \).

2. **Apply the tangent addition formula:**
We can use the tangent addition formula:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]
Since \( \alpha + \beta = \frac{\pi}{4} \), we have:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]
Therefore, we can set up the equation:
\[ 1 = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]

3. **Cross-multiply to eliminate the fraction:**
This gives us:
\[ 1 - \tan \alpha \tan \beta = \tan \alpha + \tan \beta \]

4. **Rearranging the equation:**
Rearranging the equation, we have:
\[ \tan \alpha + \tan \beta + \tan \alpha \tan \beta = 1 \]

5. **Let \( x = \tan \alpha \) and \( y = \tan \beta \):**
We can rewrite the equation as:
\[ x + y + xy = 1 \]

6. **Express \( (1 + x)(1 + y) \):**
Now, we need to find \( (1 + \tan \alpha)(1 + \tan \beta) \):
\[ (1 + x)(1 + y) = 1 + x + y + xy \]

7. **Substituting the value from the previous step:**
From the equation \( x + y + xy = 1 \), we substitute:
\[ (1 + x)(1 + y) = 1 + 1 = 2 \]

8. **Final result:**
Therefore, the value of \( (1 + \tan \alpha)(1 + \tan \beta) \) is:
\[ \boxed{2} \]