An integral solution of the equation
`tan^(-1)x+tan^(-1)(1//y)=tan^(-1)3` is

Options

A. (1, 4)
B. (4, 13)
C. (2, 1)
D. None of these

Answer

To solve the equation \( \tan^{-1} x + \tan^{-1} \left( \frac{1}{y} \right) = \tan^{-1} 3 \) for integral solutions, we can use the formula for the sum of inverse tangents:

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

### Step 1: Apply the formula
Let \( a = x \) and \( b = \frac{1}{y} \). According to the formula, we have:

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

### Step 2: Set the left-hand side equal to the right-hand side
Now, we can set this equal to \( \tan^{-1} 3 \):

\[ \tan^{-1} \left( \frac{x + \frac{1}{y}}{1 - \frac{x}{y}} \right) = \tan^{-1} 3 \]

### Step 3: Remove the inverse tangent
Since the tangent function is one-to-one, we can equate the arguments:

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

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:

\[ x + \frac{1}{y} = 3 \left( 1 - \frac{x}{y} \right) \]

### Step 5: Expand the equation
Expanding the right-hand side:

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

### Step 6: Rearrange the equation
Rearranging terms leads to:

\[ x + \frac{3x}{y} + \frac{1}{y} - 3 = 0 \]

### Step 7: Combine terms
Combining the terms gives us:

\[ x + \frac{3x + 1}{y} - 3 = 0 \]

### Step 8: Isolate \( y \)
Now, isolate \( y \):

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

### Step 9: Solve for \( y \)
Thus, we can express \( y \) in terms of \( x \):

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

### Step 10: Find integral solutions
To find integral solutions, we need \( y \) to be an integer. This means \( 3 - x \) must divide \( 3x + 1 \) evenly. We can test integer values for \( x \).

### Step 11: Testing integer values
1. **For \( x = 0 \)**:
\[ y = \frac{3(0) + 1}{3 - 0} = \frac{1}{3} \quad \text{(not an integer)} \]

2. **For \( x = 1 \)**:
\[ y = \frac{3(1) + 1}{3 - 1} = \frac{4}{2} = 2 \quad \text{(integer solution: (1, 2))} \]

3. **For \( x = 2 \)**:
\[ y = \frac{3(2) + 1}{3 - 2} = \frac{7}{1} = 7 \quad \text{(integer solution: (2, 7))} \]

4. **For \( x = 3 \)**:
\[ y = \frac{3(3) + 1}{3 - 3} \quad \text{(undefined)} \]

5. **For \( x = 4 \)**:
\[ y = \frac{3(4) + 1}{3 - 4} = \frac{13}{-1} = -13 \quad \text{(integer solution: (4, -13))} \]

### Conclusion
The integral solutions of the equation are \( (1, 2) \), \( (2, 7) \), and \( (4, -13) \).