If `: cot theta + cot((pi)/4 + theta) = 2`, then : `theta` =

Options

A. `2 n pi pm(pi)/(6)`
B. `2n pi pm (pi)/3`
C. `npi pm (pi)/3`
D. `npi pm (pi)/6`

Answer

To solve the equation \( \cot \theta + \cot \left( \frac{\pi}{4} + \theta \right) = 2 \), we will follow these steps:

### Step 1: Use the cotangent addition formula
We know that:
\[ \cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B} \]
In our case, let \( A = \frac{\pi}{4} \) and \( B = \theta \). Therefore, we can write:
\[ \cot\left(\frac{\pi}{4} + \theta\right) = \frac{\cot\frac{\pi}{4} \cot \theta - 1}{\cot\frac{\pi}{4} + \cot \theta} \]

### Step 2: Substitute \( \cot \frac{\pi}{4} \)
Since \( \cot \frac{\pi}{4} = 1 \), we can substitute this into our equation:
\[ \cot\left(\frac{\pi}{4} + \theta\right) = \frac{1 \cdot \cot \theta - 1}{1 + \cot \theta} = \frac{\cot \theta - 1}{1 + \cot \theta} \]

### Step 3: Substitute into the original equation
Now substituting back into the original equation:
\[ \cot \theta + \frac{\cot \theta - 1}{1 + \cot \theta} = 2 \]

### Step 4: Simplify the equation
Let \( x = \cot \theta \). The equation becomes:
\[ x + \frac{x - 1}{1 + x} = 2 \]
To eliminate the fraction, multiply through by \( 1 + x \):
\[ x(1 + x) + (x - 1) = 2(1 + x) \]
This simplifies to:
\[ x + x^2 + x - 1 = 2 + 2x \]
Combining like terms gives:
\[ x^2 - 2x - 3 = 0 \]

### Step 5: Factor the quadratic equation
Now we can factor the quadratic:
\[ (x - 3)(x + 1) = 0 \]
Thus, we have:
\[ x - 3 = 0 \quad \text{or} \quad x + 1 = 0 \]
This gives us:
\[ x = 3 \quad \text{or} \quad x = -1 \]

### Step 6: Solve for \( \theta \)
Recall that \( x = \cot \theta \):
1. For \( \cot \theta = 3 \):
\[ \theta = \cot^{-1}(3) \]
2. For \( \cot \theta = -1 \):
\[ \theta = \cot^{-1}(-1) = \frac{3\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]

### Final Result
Thus, the general solutions for \( \theta \) are:
\[ \theta = \cot^{-1}(3) + n\pi \quad \text{and} \quad \theta = \frac{3\pi}{4} + n\pi \]