For the equation `1-2x-x^2=tan^2(x+y)+cot^2(x+y)` exactly one value of `x` exists exactly two values of `x` exists `y=-1+npi+pi/4,n in Z` `y=1+npi+pi/4, n in Z`

Options

A. exactly one values of x exists
B. exactly two values of x exists
C. `y=-1+n pi+pi//4, n in Z`
D. `y=1+npi+pi//4, n in Z`

Explanation

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

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

### Step 2: Simplify the left-hand side
Factor the left-hand side:
\[ 1 - 2x - x^2 = -(x^2 + 2x - 1) \]
This can be rewritten as:
\[ -(x + 1)^2 + 1 \]
Thus, the left-hand side becomes:
\[ -(x + 1)^2 \]

### Step 3: Simplify the right-hand side
Recall that:
\[ \tan^2(x+y) + \cot^2(x+y) = \frac{\sin^2(x+y)}{\cos^2(x+y)} + \frac{\cos^2(x+y)}{\sin^2(x+y)} \]
This can be simplified using the identity:
\[ \tan^2 \theta + \cot^2 \theta = \frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta \cos^2 \theta} \]
This expression is always greater than or equal to 2, since:
\[ \tan^2 \theta + \cot^2 \theta \geq 2 \]

### Step 4: Set the equation
Now we have:
\[ -(x + 1)^2 = \tan^2(x+y) + \cot^2(x+y) \]
Since the left-hand side is less than or equal to 0 and the right-hand side is greater than or equal to 2, we can conclude:
\[ -(x + 1)^2 \leq 0 \quad \text{and} \quad \tan^2(x+y) + \cot^2(x+y) \geq 2 \]
This means that the equation can only hold true when both sides are equal to 0.

### Step 5: Solve for \(x\)
For the left-hand side to equal 0:
\[ -(x + 1)^2 = 0 \implies (x + 1)^2 = 0 \implies x + 1 = 0 \implies x = -1 \]

### Step 6: Solve for \(y\)
Now substitute \(x = -1\) into the right-hand side:
\[ \tan^2(-1 + y) + \cot^2(-1 + y) = 2 \]
This implies:
\[ \tan^2(-1 + y) = 1 \implies -1 + y = n\pi + \frac{\pi}{4} \quad \text{or} \quad -1 + y = n\pi - \frac{\pi}{4} \]
Thus, we can express \(y\) as:
\[ y = n\pi + 1 + \frac{\pi}{4} \quad \text{or} \quad y = n\pi + 1 - \frac{\pi}{4} \]

### Conclusion
From the above, we conclude:
- There is exactly one value of \(x\) which is \(x = -1\).
- The values of \(y\) can be expressed as \(y = 1 + n\pi + \frac{\pi}{4}\) or \(y = 1 + n\pi - \frac{\pi}{4}\), where \(n \in \mathbb{Z}\).

### Final Answer
- Exactly one value of \(x\) exists: \(x = -1\).
- \(y = 1 + n\pi + \frac{\pi}{4}, n \in \mathbb{Z}\) is one of the valid forms for \(y\).