`tan^(-1)((1)/(2))+tan^(-1)((1)/(3))=`

Options

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

Answer

To solve the problem \( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) \), we will use the formula for the sum of two inverse tangents:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]

provided that \( ab < 1 \).

### Step 1: Identify \( a \) and \( b \)
Let \( a = \frac{1}{2} \) and \( b = \frac{1}{3} \).

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

### Step 3: Apply the formula
Now, substitute \( a \) and \( b \) into the formula:
\[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}}\right) \]

### Step 4: Simplify the numerator
Calculate the numerator:
\[ \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]

### Step 5: Simplify the denominator
Calculate the denominator:
\[ 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6} \]

### Step 6: Combine the results
Now, substitute back into the formula:
\[ \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}(1) \]

### Step 7: Find the final result
We know that:
\[ \tan^{-1}(1) = \frac{\pi}{4} \]

Thus, the final answer is:
\[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{4} \]

### Final Answer:
\[ \frac{\pi}{4} \]