If `a^2+2a+cosec^2(pi/2(a+x))=0`, then, find the values of a and x.

Explanation

To solve the equation \( a^2 + 2a + \csc^2\left(\frac{\pi}{2}(a+x)\right) = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ a^2 + 2a + \csc^2\left(\frac{\pi}{2}(a+x)\right) = 0 \]
We can add and subtract 1 to the equation:
\[ a^2 + 2a + 1 + \csc^2\left(\frac{\pi}{2}(a+x)\right) - 1 = 0 \]
This simplifies to:
\[ (a + 1)^2 + \csc^2\left(\frac{\pi}{2}(a+x)\right) - 1 = 0 \]

### Step 2: Use the identity for cosecant
Recall that \(\csc^2\theta - 1 = \cot^2\theta\). Therefore, we can rewrite the equation as:
\[ (a + 1)^2 + \cot^2\left(\frac{\pi}{2}(a+x)\right) = 0 \]

### Step 3: Analyze the equation
Both \( (a + 1)^2 \) and \( \cot^2\left(\frac{\pi}{2}(a+x)\right) \) are non-negative (since squares are always non-negative). For their sum to equal zero, both terms must individually equal zero:
1. \( (a + 1)^2 = 0 \)
2. \( \cot^2\left(\frac{\pi}{2}(a+x)\right) = 0 \)

### Step 4: Solve for \( a \)
From \( (a + 1)^2 = 0 \), we find:
\[ a + 1 = 0 \implies a = -1 \]

### Step 5: Solve for \( x \)
Now, we solve \( \cot^2\left(\frac{\pi}{2}(a+x)\right) = 0 \). The cotangent function is zero when its argument is an odd multiple of \(\frac{\pi}{2}\):
\[ \frac{\pi}{2}(a + x) = \frac{\pi}{2}(2n + 1) \quad \text{for } n \in \mathbb{Z} \]
Substituting \( a = -1 \):
\[ \frac{\pi}{2}(-1 + x) = \frac{\pi}{2}(2n + 1) \]
This simplifies to:
\[ -1 + x = 2n + 1 \]
Thus,
\[ x = 2n + 2 \]

### Final Values
The values of \( a \) and \( x \) are:
\[ a = -1, \quad x = 2n + 2 \quad \text{where } n \text{ is any integer.} \]