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 elements of sets A and B based on the given conditions.

### Step 1: Determine Set A
Set A is defined as:
\[ A = \{ x : \tan x = \sec x, x \in [0, 4\pi] \} \]

We know that:
\[ \tan x = \frac{\sin x}{\cos x} \]
\[ \sec x = \frac{1}{\cos x} \]

Setting these equal gives:
\[ \frac{\sin x}{\cos x} = \frac{1}{\cos x} \]

### Step 2: Simplify the Equation
Multiplying both sides by \(\cos x\) (assuming \(\cos x \neq 0\)):
\[ \sin x = 1 \]

### Step 3: Solve for x
The sine function equals 1 at:
\[ x = \frac{\pi}{2} + 2k\pi \]
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 elements of set A are:
\[ A = \left\{ \frac{\pi}{2}, \frac{5\pi}{2} \right\} \]

### Step 4: Determine Set B
Set B is defined as:
\[ B = \{ x : \sin^2 x = 1, x \in [0, 4\pi] \} \]

### Step 5: Solve for x in Set B
The equation \(\sin^2 x = 1\) can be rewritten as:
\[ \sin x = \pm 1 \]

This gives us two cases:
1. \(\sin x = 1\)
2. \(\sin x = -1\)

#### Case 1: \(\sin x = 1\)
This occurs at:
\[ x = \frac{\pi}{2} + 2k\pi \]
Within \([0, 4\pi]\), we have:
- \(x = \frac{\pi}{2}\)
- \(x = \frac{5\pi}{2}\)

#### Case 2: \(\sin x = -1\)
This occurs at:
\[ x = \frac{3\pi}{2} + 2k\pi \]
Within \([0, 4\pi]\), we have:
- \(x = \frac{3\pi}{2}\)
- \(x = \frac{7\pi}{2}\)

### Step 6: Combine Elements of Set B
Thus, the elements of set B are:
\[ B = \left\{ \frac{\pi}{2}, \frac{5\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{2} \right\} \]

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

### Conclusion
Set A is a subset of Set B:
\[ A \subset B \]