`tan^(-1)1+tan^(-1)1/3=?`

Options

A. `tan^(-1)4/3`
B. `tan^(-1)2/3`
C. `tan^(-1)2`
D. `tan^(-1)3`

Explanation

To solve the problem \( \tan^{-1}(1) + \tan^{-1}\left(\frac{1}{3}\right) \), we can use the formula for the sum of two inverse tangent functions:

\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \]

provided that \( xy < 1 \).

### Step-by-Step Solution:

**Step 1: Identify the values of \( x \) and \( y \)**
Here, we have:
- \( x = 1 \)
- \( y = \frac{1}{3} \)

**Step 2: Check the condition \( xy < 1 \)**
Calculate \( xy \):
\[ xy = 1 \cdot \frac{1}{3} = \frac{1}{3} \]
Since \( \frac{1}{3} < 1 \), we can apply the formula.

**Step 3: Apply the formula**
Using the formula:
\[ \tan^{-1}(1) + \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{1 + \frac{1}{3}}{1 - 1 \cdot \frac{1}{3}}\right) \]

**Step 4: Simplify the expression inside the inverse tangent**
Calculate the numerator:
\[ 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} \]

Calculate the denominator:
\[ 1 - 1 \cdot \frac{1}{3} = 1 - \frac{1}{3} = \frac{2}{3} \]

So, we have:
\[ \tan^{-1}\left(\frac{\frac{4}{3}}{\frac{2}{3}}\right) = \tan^{-1}\left(\frac{4}{3} \cdot \frac{3}{2}\right) = \tan^{-1}\left(2\right) \]

**Step 5: Final result**
Thus, we conclude:
\[ \tan^{-1}(1) + \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}(2) \]

### Final Answer:
\[ \tan^{-1}(1) + \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}(2) \]

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