Find all number of pairs x,y that satisfy the equation `tan^(4) x + tan^(4)y+2 cot^(2)x * cot^(2) y=3+ sin^(2)(x+y)` .

Answer

To solve the equation \( \tan^4 x + \tan^4 y + 2 \cot^2 x \cot^2 y = 3 + \sin^2(x+y) \), we will follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the equation using the identities for tangent and cotangent:
\[ \tan^4 x + \tan^4 y + 2 \cot^2 x \cot^2 y = 3 + \sin^2(x+y) \]
We know that \( \cot^2 x = \frac{1}{\tan^2 x} \) and \( \cot^2 y = \frac{1}{\tan^2 y} \). Thus, we can express \( 2 \cot^2 x \cot^2 y \) as:
\[ 2 \cot^2 x \cot^2 y = 2 \cdot \frac{1}{\tan^2 x} \cdot \frac{1}{\tan^2 y} = \frac{2}{\tan^2 x \tan^2 y} \]
Now, substituting this back into the original equation gives us:
\[ \tan^4 x + \tan^4 y + \frac{2}{\tan^2 x \tan^2 y} = 3 + \sin^2(x+y) \]

### Step 2: Use trigonometric identities
Next, we can use the identity for \( \sin^2(x+y) \):
\[ \sin^2(x+y) = \sin^2 x \cos^2 y + \cos^2 x \sin^2 y + 2 \sin x \cos x \sin y \cos y \]
However, for simplicity, we will analyze the equation without expanding this identity.

### Step 3: Analyze the equation
We can analyze the left-hand side of the equation. The expression \( \tan^4 x + \tan^4 y \) can be rewritten using the identity:
\[ a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2 \]
Let \( a = \tan^2 x \) and \( b = \tan^2 y \):
\[ \tan^4 x + \tan^4 y = (\tan^2 x + \tan^2 y)^2 - 2 \tan^2 x \tan^2 y \]

### Step 4: Set conditions for equality
Now, we can set conditions for the equality to hold. We know that \( \sin^2(x+y) \) is always between 0 and 1. Therefore, we can analyze the boundaries:
1. The left-hand side must also be non-negative.
2. The maximum value of \( \sin^2(x+y) \) is 1, which means:
\[ \tan^4 x + \tan^4 y + 2 \cot^2 x \cot^2 y \leq 4 \]

### Step 5: Solve for specific angles
To find specific solutions, we can set \( \tan^2 x = \tan^2 y \). This leads us to:
\[ \tan^2 x = \tan^2 y \implies x = y + n\pi \text{ or } x = -y + n\pi \]
for some integer \( n \).

### Step 6: Find pairs (x, y)
From the earlier analysis, we can find pairs:
1. If \( x = y \), then \( \tan^2 x = 1 \) gives \( x = \frac{\pi}{4} + k\pi \) for integers \( k \).
2. If \( x = -y \), then \( \tan^2 x = 1 \) gives \( x = \frac{\pi}{4} + k\pi \) and \( y = -\left(\frac{\pi}{4} + k\pi\right) \).

### Conclusion
Thus, the pairs \( (x, y) \) that satisfy the equation are:
\[ (x, y) = \left(\frac{\pi}{4} + k\pi, \frac{\pi}{4} + k\pi\right) \quad \text{and} \quad (x, y) = \left(\frac{\pi}{4} + k\pi, -\left(\frac{\pi}{4} + k\pi\right)\right) \]
for integers \( k \).