If tan `A=(a)/(a+1)` and tan `B=(1)/(2a+1)` then find the value of A+B

Answer

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

### Step-by-Step Solution:

1. **Use the formula for \( \tan(A + B) \)**:
The formula for \( \tan(A + B) \) is given by:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

2. **Substitute the values of \( \tan A \) and \( \tan B \)**:
Substitute \( \tan A = \frac{a}{a+1} \) and \( \tan B = \frac{1}{2a+1} \) into the formula:
\[ \tan(A + B) = \frac{\frac{a}{a+1} + \frac{1}{2a+1}}{1 - \left(\frac{a}{a+1}\right)\left(\frac{1}{2a+1}\right)} \]

3. **Find a common denominator for the numerator**:
The common denominator for the fractions in the numerator is \((a+1)(2a+1)\):
\[ \tan(A + B) = \frac{\frac{a(2a+1) + 1(a+1)}{(a+1)(2a+1)}}{1 - \frac{a}{(a+1)(2a+1)}} \]

4. **Simplify the numerator**:
Expanding the numerator:
\[ a(2a + 1) + (a + 1) = 2a^2 + a + a + 1 = 2a^2 + 2a + 1 \]
So, the numerator becomes:
\[ \frac{2a^2 + 2a + 1}{(a+1)(2a+1)} \]

5. **Simplify the denominator**:
The denominator simplifies as follows:
\[ 1 - \frac{a}{(a+1)(2a+1)} = \frac{(a+1)(2a+1) - a}{(a+1)(2a+1)} = \frac{2a^2 + 2a + 1 - a}{(a+1)(2a+1)} = \frac{2a^2 + a + 1}{(a+1)(2a+1)} \]

6. **Combine the results**:
Now substituting back into the formula for \( \tan(A + B) \):
\[ \tan(A + B) = \frac{\frac{2a^2 + 2a + 1}{(a+1)(2a+1)}}{\frac{2a^2 + a + 1}{(a+1)(2a+1)}} \]
This simplifies to:
\[ \tan(A + B) = \frac{2a^2 + 2a + 1}{2a^2 + a + 1} \]

7. **Evaluate the limit**:
To find \( A + B \), we need to find when \( \tan(A + B) = 1 \):
\[ \frac{2a^2 + 2a + 1}{2a^2 + a + 1} = 1 \]
This leads to:
\[ 2a^2 + 2a + 1 = 2a^2 + a + 1 \]
Simplifying gives:
\[ 2a = a \implies a = 0 \]

8. **Find \( A + B \)**:
Since \( \tan(A + B) = 1 \), we have:
\[ A + B = \tan^{-1}(1) = \frac{\pi}{4} \]

### Final Answer:
\[ A + B = \frac{\pi}{4} \]