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

Options

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

Explanation

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

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

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

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

Now we can use the formula for the tangent of the sum of two angles:
\[ \tan(2A + B) = \frac{\tan(2A) + \tan B}{1 - \tan(2A) \tan B} \]
Substituting \( \tan(2A) = \frac{4}{3} \) and \( \tan B = \frac{1}{3} \):
\[ \tan(2A + B) = \frac{\frac{4}{3} + \frac{1}{3}}{1 - \left(\frac{4}{3} \cdot \frac{1}{3}\right)} \]
Calculating the numerator:
\[ \tan(2A + B) = \frac{\frac{4 + 1}{3}}{1 - \frac{4}{9}} = \frac{\frac{5}{3}}{1 - \frac{4}{9}} \]
Calculating the denominator:
\[ 1 - \frac{4}{9} = \frac{9 - 4}{9} = \frac{5}{9} \]
Thus, we have:
\[ \tan(2A + B) = \frac{\frac{5}{3}}{\frac{5}{9}} \]

### Step 3: Simplify the expression

To simplify:
\[ \tan(2A + B) = \frac{5}{3} \cdot \frac{9}{5} = \frac{9}{3} = 3 \]

### Final Answer
Therefore, \( \tan(2A + B) = 3 \).

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