For which values of x between `0and2pi`, then `2cosec2x cot x-cot^(2)x=1` is true :

Options

A. 0
B. 2
C. 1
D. for all x

Answer

To solve the equation \( 2 \csc(2x) \cot(x) - \cot^2(x) = 1 \) for values of \( x \) between \( 0 \) and \( 2\pi \), we will follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that:
- \( \csc(2x) = \frac{1}{\sin(2x)} = \frac{1}{2\sin(x)\cos(x)} \)
- \( \cot(x) = \frac{\cos(x)}{\sin(x)} \)

Substituting these identities into the equation gives:
\[ 2 \cdot \frac{1}{2\sin(x)\cos(x)} \cdot \frac{\cos(x)}{\sin(x)} - \left(\frac{\cos(x)}{\sin(x)}\right)^2 = 1 \]

### Step 2: Simplify the left-hand side
This simplifies to:
\[ \frac{1}{\sin^2(x)} - \frac{\cos^2(x)}{\sin^2(x)} = 1 \]
Combining the fractions:
\[ \frac{1 - \cos^2(x)}{\sin^2(x)} = 1 \]

### Step 3: Use the Pythagorean identity
Using the identity \( 1 - \cos^2(x) = \sin^2(x) \):
\[ \frac{\sin^2(x)}{\sin^2(x)} = 1 \]
This simplifies to:
\[ 1 = 1 \]

### Step 4: Conclusion
Since the equation \( 1 = 1 \) is always true, it implies that the original equation holds for all values of \( x \) in the interval \( [0, 2\pi] \).

### Final Answer
Thus, the equation \( 2 \csc(2x) \cot(x) - \cot^2(x) = 1 \) is true for all values of \( x \) in the interval \( [0, 2\pi] \).

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