Solve for x and y `, 1-2x-x^(2)=tan^(2)(x+y)+cot^(2)(x+y)`.

Answer

To solve the equation \(1 - 2x - x^2 = \tan^2(x+y) + \cot^2(x+y)\), we will follow these steps:

### Step 1: Analyze the left side of the equation
The left side of the equation is a quadratic expression in terms of \(x\):
\[ 1 - 2x - x^2 = 0 \]
This can be rewritten as:
\[ -x^2 - 2x + 1 = 0 \]

### Step 2: Find the discriminant
To determine the nature of the roots, we calculate the discriminant \(D\):
\[ D = b^2 - 4ac = (-2)^2 - 4(-1)(1) = 4 + 4 = 8 \]
Since \(D > 0\), the quadratic has two real and distinct roots.

### Step 3: Find the range of the quadratic expression
The quadratic opens downwards (since the coefficient of \(x^2\) is negative). The maximum value occurs at the vertex, given by the formula \(x = -\frac{b}{2a}\):
\[ x = -\frac{-2}{2 \cdot -1} = -1 \]
Substituting \(x = -1\) into the quadratic:
\[ 1 - 2(-1) - (-1)^2 = 1 + 2 - 1 = 2 \]
Thus, the maximum value of the quadratic is \(2\), and since it opens downwards, the range is:
\[ (-\infty, 2] \]

### Step 4: Analyze the right side of the equation
Now consider the right side:
\[ \tan^2(x+y) + \cot^2(x+y) \]
Using the identity \(\tan^2 \theta + \cot^2 \theta \geq 2\) (AM-GM inequality), we find:
\[ \tan^2(x+y) + \cot^2(x+y) \geq 2 \]

### Step 5: Set the two sides equal
Since the left side is less than or equal to \(2\) and the right side is greater than or equal to \(2\), the only possibility for equality is:
\[ 1 - 2x - x^2 = 2 \]
This simplifies to:
\[ -x^2 - 2x + 1 - 2 = 0 \implies -x^2 - 2x - 1 = 0 \implies x^2 + 2x + 1 = 0 \]

### Step 6: Solve for \(x\)
Factoring the quadratic:
\[ (x + 1)^2 = 0 \implies x = -1 \]

### Step 7: Solve for \(y\)
Now substituting \(x = -1\) into the right side:
\[ \tan^2(-1 + y) + \cot^2(-1 + y) = 2 \]
This implies:
\[ \tan(-1 + y) = 1 \quad \text{and} \quad \cot(-1 + y) = 1 \]
Thus:
\[ -x + y = \frac{\pi}{4} + n\pi \quad \text{or} \quad -x + y = -\frac{\pi}{4} + n\pi \]
Substituting \(x = -1\):
\[ -(-1) + y = \frac{\pi}{4} + n\pi \implies 1 + y = \frac{\pi}{4} + n\pi \implies y = n\pi - 1 + \frac{\pi}{4} \]
And:
\[ 1 + y = -\frac{\pi}{4} + n\pi \implies y = n\pi - 1 - \frac{\pi}{4} \]

### Final Solution
Thus, the solutions for \(x\) and \(y\) are:
\[ x = -1, \quad y = n\pi - 1 + \frac{\pi}{4} \quad \text{or} \quad y = n\pi - 1 - \frac{\pi}{4} \]