If `tan alpha =m/(m+1)` and `tan beta =1/(2m+1)` , then `alpha+beta` is equal to

Options

A. `(pi)/(2)`
B. `(pi)/(3)`
C. `(pi)/(6)`
D. `(pi)/(4)`

Explanation

To solve the problem, we need to find the value of \( \alpha + \beta \) given that \( \tan \alpha = \frac{m}{m+1} \) and \( \tan \beta = \frac{1}{2m+1} \).

### Step-by-Step Solution:

1. **Use the Formula for \( \tan(\alpha + \beta) \)**:
We know that:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]

2. **Substitute the Values of \( \tan \alpha \) and \( \tan \beta \)**:
Substitute \( \tan \alpha = \frac{m}{m+1} \) and \( \tan \beta = \frac{1}{2m+1} \):
\[ \tan(\alpha + \beta) = \frac{\frac{m}{m+1} + \frac{1}{2m+1}}{1 - \left(\frac{m}{m+1} \cdot \frac{1}{2m+1}\right)} \]

3. **Simplify the Numerator**:
To add the fractions in the numerator:
\[ \tan(\alpha + \beta) = \frac{\frac{m(2m+1) + 1(m+1)}{(m+1)(2m+1)}}{1 - \frac{m}{(m+1)(2m+1)}} \]
Simplifying the numerator:
\[ = \frac{(2m^2 + m + m + 1)}{(m+1)(2m+1)} = \frac{2m^2 + 2m + 1}{(m+1)(2m+1)} \]

4. **Simplify the Denominator**:
The denominator becomes:
\[ 1 - \frac{m}{(m+1)(2m+1)} = \frac{(m+1)(2m+1) - m}{(m+1)(2m+1)} = \frac{2m^2 + 2m + 1 - m}{(m+1)(2m+1)} = \frac{2m^2 + m + 1}{(m+1)(2m+1)} \]

5. **Combine the Results**:
Now substituting back into the formula:
\[ \tan(\alpha + \beta) = \frac{2m^2 + 2m + 1}{2m^2 + m + 1} \]

6. **Check for Simplification**:
Notice that the numerator and denominator are equal:
\[ \tan(\alpha + \beta) = 1 \]

7. **Determine the Angle**:
We know that:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]
Therefore, we conclude:
\[ \alpha + \beta = \frac{\pi}{4} \]

### Final Answer:
\[ \alpha + \beta = \frac{\pi}{4} \]