`tan^(-1)(1/2)+tan^(-1)(1/3)` is equal to

Options

A. `pi/6`
B. `pi/4`
C. `pi/3`
D. `(5pi)/12`

Answer

To solve the expression \( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) \), 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 1: Identify \( x \) and \( y \)

Here, we have:
- \( x = \frac{1}{2} \)
- \( y = \frac{1}{3} \)

### Step 2: Check the condition

First, we need to check the condition \( xy < 1 \):
\[ xy = \left(\frac{1}{2}\right) \left(\frac{1}{3}\right) = \frac{1}{6} < 1 \]
Since this condition is satisfied, we can proceed.

### Step 3: Apply the formula

Now we apply the formula:
\[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{\frac{1}{2} + \frac{1}{3}}{1 - \left(\frac{1}{2} \cdot \frac{1}{3}\right)}\right) \]

### Step 4: Calculate the numerator

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

### Step 5: Calculate the denominator

Now calculate the denominator:
\[ 1 - \left(\frac{1}{2} \cdot \frac{1}{3}\right) = 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6} \]

### Step 6: Combine the results

Now we can combine the results:
\[ \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}(1) \]

### Step 7: Find the final value

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

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