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

Options

A. `pi/2`
B. `pi/4`
C. 0
D. `pi`

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 tangents:

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

provided that \( xy < 1 \).

### Step 1: Identify \( x \) and \( y \)

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

### Step 2: Calculate \( x + y \)

\[ x + y = \frac{1}{2} + \frac{1}{3} \]

To add these fractions, we need a common denominator. The least common multiple of 2 and 3 is 6.

\[ x + y = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]

### Step 3: Calculate \( xy \)

\[ xy = \left( \frac{1}{2} \right) \left( \frac{1}{3} \right) = \frac{1}{6} \]

### Step 4: Check the condition \( xy < 1 \)

Since \( \frac{1}{6} < 1 \), we can use the formula.

### Step 5: Substitute into the formula

Now we substitute \( x + y \) and \( xy \) into the formula:

\[ \tan^{-1} \left( \frac{x + y}{1 - xy} \right) = \tan^{-1} \left( \frac{\frac{5}{6}}{1 - \frac{1}{6}} \right) \]

### Step 6: Simplify \( 1 - xy \)

\[ 1 - xy = 1 - \frac{1}{6} = \frac{5}{6} \]

### Step 7: Substitute back into the formula

Now we have:

\[ \tan^{-1} \left( \frac{\frac{5}{6}}{\frac{5}{6}} \right) = \tan^{-1}(1) \]

### Step 8: Find the value of \( \tan^{-1}(1) \)

The value of \( \tan^{-1}(1) \) is:

\[ \tan^{-1}(1) = \frac{\pi}{4} \]

### Final Answer

Thus,

\[ \tan^{-1} \left( \frac{1}{2} \right) + \tan^{-1} \left( \frac{1}{3} \right) = \frac{\pi}{4} \]

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