Find the number of positive integral solution of the equation `Tan^(-1)x+Tan^(-1)(1//y)=Tan^(-1)3` is

Explanation

To find the number of positive integral solutions of the equation

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

we can follow these steps:

### Step 1: Rewrite the equation using the tangent subtraction formula
We can use the formula for the difference of two inverse tangents:

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

Rearranging the original equation, we have:

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

Applying the formula gives us:

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

### Step 2: Solve for \(y\)
Taking the reciprocal of both sides, we get:

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

### Step 3: Determine conditions for \(y\) to be a positive integer
For \(y\) to be a positive integer, both the numerator and denominator must be positive.

1. **Numerator**: \(1 + 3x > 0\) is always true for positive \(x\).
2. **Denominator**: \(3 - x > 0\) implies \(x < 3\).

Thus, \(x\) can take positive integer values \(1\) and \(2\).

### Step 4: Calculate \(y\) for valid values of \(x\)
- For \(x = 1\):
\[ y = \frac{1 + 3(1)}{3 - 1} = \frac{4}{2} = 2. \]
So, we have the pair \((1, 2)\).

- For \(x = 2\):
\[ y = \frac{1 + 3(2)}{3 - 2} = \frac{7}{1} = 7. \]
So, we have the pair \((2, 7)\).

### Step 5: Check for larger values of \(x\)
For \(x = 3\):
\[ y = \frac{1 + 3(3)}{3 - 3} = \frac{10}{0}, \]
which is undefined.

For \(x > 3\), the denominator \(3 - x\) becomes negative, making \(y\) negative, which is not acceptable.

### Conclusion
The only valid pairs of positive integral solutions are \((1, 2)\) and \((2, 7)\).

Thus, the number of positive integral solutions is:

\[ \boxed{2}. \]