The general solution of the equation ` tan^(2)(x + y) + cot^(2) ( x+ y) = 1 - 2x - x^(2) ` lie on the line is :

Options

A. ` x = -1 `
B. ` x = -2 `
C. ` y = -1`
D. `y = -2 `

Answer

To solve the equation \( \tan^2(x + y) + \cot^2(x + y) = 1 - 2x - x^2 \) and find the general solution that lies on a specific line, we can follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ \tan^2(x + y) + \cot^2(x + y) = 1 - 2x - x^2 \]

### Step 2: Simplify the Left-Hand Side
Recall that \( \cot^2(x + y) = \frac{1}{\tan^2(x + y)} \). Therefore, we can rewrite the left-hand side:
\[ \tan^2(x + y) + \frac{1}{\tan^2(x + y)} \]
Let \( t = \tan^2(x + y) \). Then the left-hand side becomes:
\[ t + \frac{1}{t} \]

### Step 3: Set Up the Equation
Now, we can rewrite the equation as:
\[ t + \frac{1}{t} = 1 - 2x - x^2 \]

### Step 4: Rearrange the Equation
We multiply both sides by \( t \) (assuming \( t \neq 0 \)):
\[ t^2 + 1 = (1 - 2x - x^2)t \]
Rearranging gives us:
\[ t^2 - (1 - 2x - x^2)t + 1 = 0 \]

### Step 5: Analyze the Quadratic Equation
This is a quadratic equation in \( t \). For \( t \) to have real solutions, the discriminant must be non-negative:
\[ D = (1 - 2x - x^2)^2 - 4 \cdot 1 \cdot 1 \geq 0 \]

### Step 6: Solve the Discriminant
Calculating the discriminant:
\[ D = (1 - 2x - x^2)^2 - 4 \]
Setting \( D = 0 \) for real solutions:
\[ (1 - 2x - x^2)^2 - 4 = 0 \]
This simplifies to:
\[ (1 - 2x - x^2)^2 = 4 \]
Taking the square root of both sides gives:
\[ 1 - 2x - x^2 = 2 \quad \text{or} \quad 1 - 2x - x^2 = -2 \]

### Step 7: Solve Each Case
**Case 1:**
\[ 1 - 2x - x^2 = 2 \implies -x^2 - 2x - 1 = 0 \implies x^2 + 2x + 1 = 0 \implies (x + 1)^2 = 0 \implies x = -1 \]

**Case 2:**
\[ 1 - 2x - x^2 = -2 \implies -x^2 - 2x + 3 = 0 \implies x^2 + 2x - 3 = 0 \]
Factoring gives:
\[ (x + 3)(x - 1) = 0 \implies x = -3 \text{ or } x = 1 \]

### Step 8: Conclusion
The solutions for \( x \) are \( x = -1, -3, 1 \). However, since we are looking for the line on which the general solution lies, we find that the primary solution is:
\[ x = -1 \]
Thus, the general solution of the equation lies on the line \( x = -1 \).

### Final Answer
The general solution of the equation lies on the line:
\[ \boxed{x = -1} \]