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 find the value of \( \tan(2A + B) \) given \( \tan A = \frac{1}{2} \) and \( \tan B = \frac{1}{3} \), we can follow these steps:

### Step 1: Calculate \( \tan(2A) \)
We will use the formula for \( \tan(2A) \):
\[ \tan(2A) = \frac{2 \tan A}{1 - \tan^2 A} \]
Substituting \( \tan A = \frac{1}{2} \):
\[ \tan(2A) = \frac{2 \cdot \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} \]
Calculating the numerator:
\[ 2 \cdot \frac{1}{2} = 1 \]
Calculating the denominator:
\[ 1 - \left(\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4} \]
Thus,
\[ \tan(2A) = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

### Step 2: Calculate \( \tan(2A + B) \)
Now we will use the formula for \( \tan(2A + B) \):
\[ \tan(2A + B) = \frac{\tan(2A) + \tan B}{1 - \tan(2A) \tan B} \]
Substituting \( \tan(2A) = \frac{4}{3} \) and \( \tan B = \frac{1}{3} \):
\[ \tan(2A + B) = \frac{\frac{4}{3} + \frac{1}{3}}{1 - \left(\frac{4}{3}\right) \left(\frac{1}{3}\right)} \]
Calculating the numerator:
\[ \frac{4}{3} + \frac{1}{3} = \frac{5}{3} \]
Calculating the denominator:
\[ 1 - \left(\frac{4}{3} \cdot \frac{1}{3}\right) = 1 - \frac{4}{9} = \frac{5}{9} \]
Thus,
\[ \tan(2A + B) = \frac{\frac{5}{3}}{\frac{5}{9}} = \frac{5}{3} \cdot \frac{9}{5} = 3 \]

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

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