If `"tan"A=1/(2),"tan"B=1/(3), then "tan"(2A+B)` is equal to

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the value of \( \tan(2A + B) \) given that \( \tan A = \frac{1}{2} \) and \( \tan B = \frac{1}{3} \).

### Step 1: Find \( \tan(2A) \)

We use the double angle formula for tangent:
\[ \tan(2A) = \frac{2 \tan A}{1 - \tan^2 A} \]
Substituting \( \tan A = \frac{1}{2} \):
\[ \tan(2A) = \frac{2 \cdot \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

### Step 2: Find \( \tan(2A + B) \)

Now we apply the tangent addition formula:
\[ \tan(2A + B) = \frac{\tan(2A) + \tan B}{1 - \tan(2A) \tan B} \]
Substituting \( \tan(2A) = \frac{4}{3} \) and \( \tan B = \frac{1}{3} \):
\[ \tan(2A + B) = \frac{\frac{4}{3} + \frac{1}{3}}{1 - \left(\frac{4}{3} \cdot \frac{1}{3}\right)} \]

### Step 3: Simplify the numerator and denominator

Calculating the numerator:
\[ \frac{4}{3} + \frac{1}{3} = \frac{5}{3} \]

Calculating the denominator:
\[ 1 - \left(\frac{4}{3} \cdot \frac{1}{3}\right) = 1 - \frac{4}{9} = \frac{9}{9} - \frac{4}{9} = \frac{5}{9} \]

### Step 4: Combine the results

Now substituting back into the formula:
\[ \tan(2A + B) = \frac{\frac{5}{3}}{\frac{5}{9}} = \frac{5}{3} \cdot \frac{9}{5} = 3 \]

Thus, the final answer is:
\[ \boxed{3} \]