Find the number of pairs of integer `(x , y)` that satisfy the following two equations: `{cos(x y)=xtan(x y)=y` 1 (b) 2 (c) 4 (d) 6

Options

A. 1
B. 2
C. 4
D. 6

Explanation

To solve the problem of finding the number of pairs of integers \((x, y)\) that satisfy the equations \( \cos(xy) = x \) and \( \tan(xy) = y \), we will follow these steps:

### Step 1: Write down the equations
We start with the two equations:
1. \( \cos(xy) = x \) (Equation 1)
2. \( \tan(xy) = y \) (Equation 2)

### Step 2: Analyze Equation 2
From Equation 2, we know that:
\[ \tan(xy) = \frac{\sin(xy)}{\cos(xy)} = y \]
This implies:
\[ \sin(xy) = y \cdot \cos(xy) \]
Substituting Equation 1 into this gives:
\[ \sin(xy) = y \cdot x \]

### Step 3: Set up the relationship
We now have:
1. \( \cos(xy) = x \)
2. \( \sin(xy) = y \cdot x \)

### Step 4: Use the identity \( \sin^2 + \cos^2 = 1 \)
Using the Pythagorean identity:
\[ \sin^2(xy) + \cos^2(xy) = 1 \]
Substituting our expressions:
\[ (y \cdot x)^2 + x^2 = 1 \]
This simplifies to:
\[ y^2 x^2 + x^2 = 1 \]
Factoring out \(x^2\):
\[ x^2(y^2 + 1) = 1 \]

### Step 5: Solve for \(x^2\)
From the equation \(x^2(y^2 + 1) = 1\), we can express \(x^2\) as:
\[ x^2 = \frac{1}{y^2 + 1} \]
Since \(x^2\) must be non-negative, \(y^2 + 1\) must be positive, which it always is. Therefore, \(x\) can be expressed as:
\[ x = \pm \sqrt{\frac{1}{y^2 + 1}} \]

### Step 6: Determine integer solutions
For \(x\) to be an integer, \(\sqrt{\frac{1}{y^2 + 1}}\) must be a rational number. This occurs when \(y^2 + 1\) is a perfect square. Let:
\[ y^2 + 1 = k^2 \quad \text{for some integer } k \]
This implies:
\[ k^2 - y^2 = 1 \implies (k - y)(k + y) = 1 \]
The integer pairs \((k - y, k + y)\) that multiply to 1 are:
1. \((1, 1)\) leading to \(k - y = 1\) and \(k + y = 1\) which gives \(k = 1\) and \(y = 0\).
2. \((-1, -1)\) leading to \(k - y = -1\) and \(k + y = -1\) which gives \(k = -1\) and \(y = 0\).

### Step 7: Find corresponding \(x\) values
For \(y = 0\):
\[ x^2 = \frac{1}{0^2 + 1} = 1 \implies x = \pm 1 \]
Thus, we have the pairs:
1. \((1, 0)\)
2. \((-1, 0)\)

### Step 8: Check for other values of \(y\)
For \(y = 1\):
\[ x^2 = \frac{1}{1^2 + 1} = \frac{1}{2} \quad \text{(not an integer)} \]
For \(y = -1\):
\[ x^2 = \frac{1}{(-1)^2 + 1} = \frac{1}{2} \quad \text{(not an integer)} \]
For \(y = 2\) and \(y = -2\):
\[ x^2 = \frac{1}{4 + 1} = \frac{1}{5} \quad \text{(not an integer)} \]
Continuing this process shows that no other integer values for \(y\) yield integer \(x\).

### Conclusion
The only integer pairs \((x, y)\) that satisfy the equations are:
1. \((1, 0)\)
2. \((-1, 0)\)

Thus, the total number of pairs of integers \((x, y)\) is **2**.

### Final Answer
The answer is **2** (Option B).