The vlaue of `tan^(-1)(1/2)+tan^(-1)(1/3)` is

Options

A. 0
B. `(pi)/(3)`
C. `(pi)/(6)`
D. `(pi)/(4)`

Explanation

To find the value of \( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) \), 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:

1. **Identify \( x \) and \( y \)**:
Let \( x = \frac{1}{2} \) and \( y = \frac{1}{3} \).

2. **Check the condition \( xy < 1 \)**:
Calculate \( xy \):
\[ xy = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \]
Since \( \frac{1}{6} < 1 \), we can use the formula.

3. **Apply the formula**:
Substitute \( x \) and \( y \) into the formula:
\[ \tan^{-1}\left(\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}}\right) \]

4. **Calculate the numerator**:
Find \( \frac{1}{2} + \frac{1}{3} \):
\[ \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]

5. **Calculate the denominator**:
Find \( 1 - \frac{1}{6} \):
\[ 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6} \]

6. **Combine the results**:
Substitute the results back into the formula:
\[ \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}(1) \]

7. **Find the final value**:
We know that \( \tan^{-1}(1) = \frac{\pi}{4} \).

Thus, the value of \( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) \) is:

\[ \frac{\pi}{4} \]