Statement-1: The general solution of `tan 5theta = cot 2theta` is `theta = (npi)/7 + pi/14`, `n in Z`. and Statement-2: The equation `cos theta =k` has exactly two solutions in `[0, 2pi]` for all `k, -1 le k le 1`.

Options

A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for statement-1
B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct expanation for statement-1
C. Statement-1 is True, Statement-2 is false
D. Statement-1 is False, Statement-2 is true

Answer

To solve the problem, we will analyze both statements step by step.

### Step 1: Analyze Statement 1
The first statement is:
**Statement 1**: The general solution of \( \tan 5\theta = \cot 2\theta \) is \( \theta = \frac{n\pi}{7} + \frac{\pi}{14} \), where \( n \in \mathbb{Z} \).

1. Start with the equation:
\[ \tan 5\theta = \cot 2\theta \]
We know that \( \cot x = \frac{1}{\tan x} \), so we can rewrite the equation as:
\[ \tan 5\theta = \frac{1}{\tan 2\theta} \]

2. This implies:
\[ \tan 5\theta \tan 2\theta = 1 \]

3. Using the identity \( \tan A \tan B = 1 \) leads to:
\[ 5\theta + 2\theta = n\pi \quad \text{or} \quad 5\theta - 2\theta = n\pi \]
This simplifies to:
\[ 7\theta = n\pi \quad \text{or} \quad 3\theta = n\pi \]

4. Solving for \( \theta \):
- From \( 7\theta = n\pi \):
\[ \theta = \frac{n\pi}{7} \]
- From \( 3\theta = n\pi \):
\[ \theta = \frac{n\pi}{3} \]

5. However, we need to consider the cotangent function:
\[ \tan 5\theta = \tan\left(\frac{\pi}{2} - 2\theta\right) \]
This gives:
\[ 5\theta = \frac{\pi}{2} - 2\theta + k\pi \quad (k \in \mathbb{Z}) \]
Rearranging this gives:
\[ 7\theta = \frac{\pi}{2} + k\pi \]
Thus:
\[ \theta = \frac{(2k + 1)\pi}{14} \quad (k \in \mathbb{Z}) \]

6. Therefore, the general solution can be expressed as:
\[ \theta = \frac{n\pi}{7} + \frac{\pi}{14} \quad (n \in \mathbb{Z}) \]
This confirms that **Statement 1 is true**.

### Step 2: Analyze Statement 2
The second statement is:
**Statement 2**: The equation \( \cos \theta = k \) has exactly two solutions in the interval \( [0, 2\pi] \) for all \( k \) such that \( -1 \leq k \leq 1 \).

1. The cosine function is periodic with a period of \( 2\pi \) and is symmetric about the y-axis.
2. The range of \( \cos \theta \) is from -1 to 1.
3. For each value \( k \) in the interval \( [-1, 1] \):
- If \( k = 1 \), \( \cos \theta = 1 \) has one solution: \( \theta = 0 \).
- If \( k = -1 \), \( \cos \theta = -1 \) has one solution: \( \theta = \pi \).
- For \( -1 < k < 1 \), \( \cos \theta = k \) has two solutions in the interval \( [0, 2\pi] \): one in the first quadrant and one in the fourth quadrant.

4. Therefore, the statement that \( \cos \theta = k \) has exactly two solutions in \( [0, 2\pi] \) is **not true for \( k = 1 \) and \( k = -1 \)**, as it only has one solution in those cases.

### Conclusion
- **Statement 1** is true.
- **Statement 2** is false because it does not hold for \( k = 1 \) and \( k = -1 \).

### Final Answer
Statement 1 is true, and Statement 2 is false.