If `tan alpha = 1/3` and `tan beta = 1/2`, then: `tan(alpha + beta)`=

Options

A. 1
B. `-1`
C. 2
D. 3

Answer

To solve the problem, we need to find \( \tan(\alpha + \beta) \) given that \( \tan \alpha = \frac{1}{3} \) and \( \tan \beta = \frac{1}{2} \). We will 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 \( \tan \beta \)

Given:
- \( \tan \alpha = \frac{1}{3} \)
- \( \tan \beta = \frac{1}{2} \)

Substituting these values into the formula:

\[ \tan(\alpha + \beta) = \frac{\frac{1}{3} + \frac{1}{2}}{1 - \left(\frac{1}{3} \cdot \frac{1}{2}\right)} \]

### Step 2: Calculate the numerator

To add \( \frac{1}{3} \) and \( \frac{1}{2} \), we need a common denominator. The least common multiple (LCM) of 3 and 2 is 6.

\[ \frac{1}{3} = \frac{2}{6}, \quad \frac{1}{2} = \frac{3}{6} \]

Now, adding these fractions:

\[ \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6} \]

### Step 3: Calculate the denominator

Now we calculate \( 1 - \left(\frac{1}{3} \cdot \frac{1}{2}\right) \):

\[ \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} \]

Thus, the denominator becomes:

\[ 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6} \]

### Step 4: Substitute back into the formula

Now we can substitute the values of the numerator and denominator back into the formula:

\[ \tan(\alpha + \beta) = \frac{\frac{5}{6}}{\frac{5}{6}} \]

### Step 5: Simplify the expression

This simplifies to:

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

### Final Answer

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

\[ \boxed{1} \]

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