If set `A={x:tanx = secx, x in [0, 4pi]} and " set "B={x:sin^(2)x=1, x in [0, 4pi]}`, then

Options

A. `A sub B`
B. `A=B`
C. `AnnB=B`
D. `n(AxxB)=0`

Answer

To solve the problem, we need to find the sets A and B based on the given equations and then analyze their relationship.

### Step 1: Finding Set A
Set A is defined by the equation \( \tan x = \sec x \) for \( x \) in the interval \( [0, 4\pi] \).

1. **Rewrite the equation**:
\[ \tan x = \sec x \implies \frac{\sin x}{\cos x} = \frac{1}{\cos x} \]
This simplifies to:
\[ \sin x = 1 \]

2. **Solve for x**:
The solutions to \( \sin x = 1 \) occur at:
\[ x = \frac{\pi}{2} + 2k\pi \quad \text{for integers } k \]
Within the interval \( [0, 4\pi] \):
- For \( k = 0 \): \( x = \frac{\pi}{2} \)
- For \( k = 1 \): \( x = \frac{\pi}{2} + 2\pi = \frac{5\pi}{2} \)

Thus, the set A is:
\[ A = \left\{ \frac{\pi}{2}, \frac{5\pi}{2} \right\} \]

### Step 2: Finding Set B
Set B is defined by the equation \( \sin^2 x = 1 \) for \( x \) in the interval \( [0, 4\pi] \).

1. **Rewrite the equation**:
\[ \sin^2 x - 1 = 0 \implies (\sin x - 1)(\sin x + 1) = 0 \]

2. **Solve for x**:
- From \( \sin x - 1 = 0 \):
\[ \sin x = 1 \implies x = \frac{\pi}{2} + 2k\pi \quad \text{for integers } k \]
- \( k = 0 \): \( x = \frac{\pi}{2} \)
- \( k = 1 \): \( x = \frac{5\pi}{2} \)

- From \( \sin x + 1 = 0 \):
\[ \sin x = -1 \implies x = \frac{3\pi}{2} + 2k\pi \quad \text{for integers } k \]
- \( k = 0 \): \( x = \frac{3\pi}{2} \)
- \( k = 1 \): \( x = \frac{7\pi}{2} \)

Thus, the set B is:
\[ B = \left\{ \frac{\pi}{2}, \frac{5\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{2} \right\} \]

### Step 3: Analyzing the Relationship between Sets A and B
Now we compare sets A and B:
- Set A: \( \left\{ \frac{\pi}{2}, \frac{5\pi}{2} \right\} \)
- Set B: \( \left\{ \frac{\pi}{2}, \frac{5\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{2} \right\} \)

### Conclusion
- **Set A is a subset of Set B**: All elements of A are in B.
- **Set A is not equal to Set B**: A has 2 elements, while B has 4 elements.

Thus, the correct answer is that \( A \subseteq B \).