If `tanalpha=m/(m+1)` and `tanbeta=1/(2m+1)`. Find the possible values of `tan(alpha+beta)`

Options

A. `2`
B. `1`
C. `-1`
D. 0

Explanation

To find the possible values of \( \tan(\alpha + \beta) \) given \( \tan \alpha = \frac{m}{m + 1} \) and \( \tan \beta = \frac{1}{2m + 1} \), we can use the formula for the tangent of the sum of two angles:

\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]

### Step 1: Substitute the values of \( \tan \alpha \) and \( \tan \beta \)

Substituting the given values into the formula:

\[ \tan(\alpha + \beta) = \frac{\frac{m}{m + 1} + \frac{1}{2m + 1}}{1 - \left(\frac{m}{m + 1} \cdot \frac{1}{2m + 1}\right)} \]

### Step 2: Find a common denominator for the numerator

The common denominator for the numerator \( \frac{m}{m + 1} + \frac{1}{2m + 1} \) is \( (m + 1)(2m + 1) \):

\[ \tan(\alpha + \beta) = \frac{\frac{m(2m + 1) + 1(m + 1)}{(m + 1)(2m + 1)}}{1 - \frac{m}{m + 1} \cdot \frac{1}{2m + 1}} \]

### Step 3: Simplify the numerator

Expanding the numerator:

\[ m(2m + 1) + (m + 1) = 2m^2 + m + m + 1 = 2m^2 + 2m + 1 \]

So, the numerator becomes:

\[ \frac{2m^2 + 2m + 1}{(m + 1)(2m + 1)} \]

### Step 4: Simplify the denominator

Now, simplify the denominator:

\[ 1 - \left(\frac{m}{m + 1} \cdot \frac{1}{2m + 1}\right) = 1 - \frac{m}{(m + 1)(2m + 1)} \]

Finding a common denominator for the denominator:

\[ = \frac{(m + 1)(2m + 1) - m}{(m + 1)(2m + 1)} = \frac{(2m^2 + 3m + 1) - m}{(m + 1)(2m + 1)} = \frac{2m^2 + 2m + 1}{(m + 1)(2m + 1)} \]

### Step 5: Combine the results

Now substituting back into the tangent formula:

\[ \tan(\alpha + \beta) = \frac{\frac{2m^2 + 2m + 1}{(m + 1)(2m + 1)}}{\frac{2m^2 + 2m + 1}{(m + 1)(2m + 1)}} \]

This simplifies to:

\[ \tan(\alpha + \beta) = 1 \]

### Conclusion

Thus, the possible value of \( \tan(\alpha + \beta) \) is:

\[ \boxed{1} \]