`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 problem \( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) \), we can use the identity for the sum of inverse tangents:

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

provided that \( ab < 1 \).

### Step-by-step Solution:

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

**Step 2: Calculate \( a + b \)**
\[ a + b = \frac{1}{2} + \frac{1}{3} \]
To add these fractions, we need a common denominator:
\[ \frac{1}{2} = \frac{3}{6}, \quad \frac{1}{3} = \frac{2}{6} \]
Thus,
\[ a + b = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]

**Step 3: Calculate \( ab \)**
\[ ab = \left(\frac{1}{2}\right) \left(\frac{1}{3}\right) = \frac{1}{6} \]

**Step 4: Check the condition \( ab < 1 \)**
Since \( \frac{1}{6} < 1 \), we can use the identity.

**Step 5: Substitute into the identity**
Now we can substitute \( a + b \) and \( ab \) into the identity:
\[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{\frac{5}{6}}{1 - \frac{1}{6}}\right) \]

**Step 6: Simplify the denominator**
Calculate \( 1 - ab \):
\[ 1 - ab = 1 - \frac{1}{6} = \frac{5}{6} \]

**Step 7: Substitute back into the equation**
Now, substituting back:
\[ \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}(1) \]

**Step 8: Find the value of \( \tan^{-1}(1) \)**
We know that:
\[ \tan^{-1}(1) = \frac{\pi}{4} \]

### Final Answer:
Thus, the value of \( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) \) is:
\[ \frac{\pi}{4} \]