If `tanalpha=(m)/(m+1)andtan beta=(1)/(2m+1)`, then `(alpha+beta)=?`

Options

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

Answer

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

### Step-by-step Solution:

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

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

3. **Calculate the numerator**:
- Find a common denominator for the numerator:
\[ \text{Numerator} = \frac{m(2m+1) + 1(m+1)}{(m+1)(2m+1)} \]
- Simplifying:
\[ = \frac{2m^2 + m + m + 1}{(m+1)(2m+1)} = \frac{2m^2 + 2m + 1}{(m+1)(2m+1)} \]

4. **Calculate the denominator**:
- The denominator becomes:
\[ 1 - \frac{m}{m+1} \cdot \frac{1}{2m+1} = 1 - \frac{m}{(m+1)(2m+1)} \]
- Finding a common denominator:
\[ = \frac{(m+1)(2m+1) - m}{(m+1)(2m+1)} = \frac{2m^2 + 3m + 1 - m}{(m+1)(2m+1)} = \frac{2m^2 + 2m + 1}{(m+1)(2m+1)} \]

5. **Combine the results**:
\[ \tan(\alpha + \beta) = \frac{\frac{2m^2 + 2m + 1}{(m+1)(2m+1)}}{\frac{2m^2 + 2m + 1}{(m+1)(2m+1)}} = 1 \]

6. **Determine \( \alpha + \beta \)**:
- Since \( \tan(\alpha + \beta) = 1 \), we know:
\[ \alpha + \beta = 45^\circ \quad \text{or} \quad \alpha + \beta = \frac{\pi}{4} \]

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