Number of ordered pairs (a, x) satisfying the equation `sec^2(a+2)x+a^2-1=0;-pi < x< pi` is

Options

A. 2
B. 1
C. 3
D. infinite

Explanation

To solve the equation \( \sec^2(a + 2x) + a^2 - 1 = 0 \) for the number of ordered pairs \((a, x)\) satisfying the condition \(-\pi < x < \pi\), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ \sec^2(a + 2x) + a^2 - 1 = 0 \]
Rearranging gives:
\[ \sec^2(a + 2x) = 1 - a^2 \]

### Step 2: Understanding the Range of \(a\)
Since \(\sec^2\) is always greater than or equal to 1, we have:
\[ 1 - a^2 \geq 1 \implies a^2 \leq 0 \]
This implies that:
\[ a^2 = 0 \implies a = 0 \]

### Step 3: Substituting \(a = 0\)
Now substituting \(a = 0\) back into the equation gives:
\[ \sec^2(2x) = 1 \]
This simplifies to:
\[ \sec^2(2x) - 1 = 0 \]
Using the identity \(\sec^2 \theta - 1 = \tan^2 \theta\), we can write:
\[ \tan^2(2x) = 0 \]

### Step 4: Solving for \(x\)
The equation \(\tan^2(2x) = 0\) implies:
\[ 2x = n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus,
\[ x = \frac{n\pi}{2} \]

### Step 5: Finding Valid Values of \(x\)
Now we need to find values of \(n\) such that \(-\pi < x < \pi\):
- For \(n = -2\): \(x = -\pi\) (not included)
- For \(n = -1\): \(x = -\frac{\pi}{2}\)
- For \(n = 0\): \(x = 0\)
- For \(n = 1\): \(x = \frac{\pi}{2}\)
- For \(n = 2\): \(x = \pi\) (not included)

Thus, the valid values of \(x\) are:
- \(x = -\frac{\pi}{2}\)
- \(x = 0\)
- \(x = \frac{\pi}{2}\)

### Step 6: Counting Ordered Pairs
Since \(a\) can only be \(0\) and we have three valid values of \(x\), the ordered pairs \((a, x)\) are:
1. \((0, -\frac{\pi}{2})\)
2. \((0, 0)\)
3. \((0, \frac{\pi}{2})\)

Thus, the total number of ordered pairs \((a, x)\) is **3**.

### Final Answer
The number of ordered pairs \((a, x)\) satisfying the equation is **3**.

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