The general solution of the equation, `2cot.(theta)/(2)=(1+cot theta)^(2)` is `(n in Z)`

Options

A. `n pi+(-1)^(n)(pi)/(4)`
B. `n pi+(-1)^(n)(pi)/(6)`
C. `n pi+(-1)^(n)(pi)/(3)`
D. none of these

Explanation

To solve the equation \( \frac{2 \cot \theta}{2} = (1 + \cot \theta)^2 \), we will follow these steps:

### Step 1: Simplify the left-hand side
The left-hand side simplifies to:
\[ \cot \theta = (1 + \cot \theta)^2 \]

### Step 2: Expand the right-hand side
Now, we expand the right-hand side:
\[ \cot \theta = 1 + 2 \cot \theta + \cot^2 \theta \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ \cot^2 \theta + 2 \cot \theta + 1 - \cot \theta = 0 \]
This simplifies to:
\[ \cot^2 \theta + \cot \theta + 1 = 0 \]

### Step 4: Solve the quadratic equation
Now, we can use the quadratic formula to solve for \( \cot \theta \):
\[ \cot \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 1, c = 1 \):
\[ \cot \theta = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ \cot \theta = \frac{-1 \pm \sqrt{1 - 4}}{2} \]
\[ \cot \theta = \frac{-1 \pm \sqrt{-3}}{2} \]
\[ \cot \theta = \frac{-1 \pm i\sqrt{3}}{2} \]

### Step 5: Find the general solution for \( \theta \)
Since \( \cot \theta \) is complex, we can express \( \theta \) in terms of the inverse cotangent:
\[ \theta = \cot^{-1}\left(\frac{-1 \pm i\sqrt{3}}{2}\right) + n\pi, \quad n \in \mathbb{Z} \]

### Final Answer
The general solution of the equation is:
\[ \theta = n\pi + \cot^{-1}\left(\frac{-1 + i\sqrt{3}}{2}\right) \quad \text{or} \quad \theta = n\pi + \cot^{-1}\left(\frac{-1 - i\sqrt{3}}{2}\right), \quad n \in \mathbb{Z} \]