If `2pi lt theta lt 4pi` , the sum of all the `theta`'s that satisfy `tan theta + 1/(tan theta) - 2=0` is ?

Options

A. `13/3 pi`
B. `13/2pi`
C. `10/3pi`
D. `11/2pi`

Answer

To solve the equation \( \tan \theta + \frac{1}{\tan \theta} - 2 = 0 \) for \( 2\pi < \theta < 4\pi \), we can follow these steps:

### Step 1: Simplify the Equation
Start with the equation:
\[ \tan \theta + \frac{1}{\tan \theta} - 2 = 0 \]
Let \( x = \tan \theta \). The equation becomes:
\[ x + \frac{1}{x} - 2 = 0 \]
Multiplying through by \( x \) (assuming \( x \neq 0 \)):
\[ x^2 - 2x + 1 = 0 \]

### Step 2: Factor the Quadratic
The equation can be factored as:
\[ (x - 1)^2 = 0 \]
This gives us:
\[ x - 1 = 0 \implies x = 1 \]
Thus, we have:
\[ \tan \theta = 1 \]

### Step 3: Find the General Solutions for \( \theta \)
The general solution for \( \tan \theta = 1 \) is:
\[ \theta = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 4: Determine Specific Solutions in the Given Range
We need to find values of \( \theta \) in the range \( 2\pi < \theta < 4\pi \).

1. For \( n = 5 \):
\[ \theta = \frac{\pi}{4} + 5\pi = \frac{\pi}{4} + \frac{20\pi}{4} = \frac{21\pi}{4} \]
(This is valid since \( 2\pi < \frac{21\pi}{4} < 4\pi \))

2. For \( n = 6 \):
\[ \theta = \frac{\pi}{4} + 6\pi = \frac{\pi}{4} + \frac{24\pi}{4} = \frac{25\pi}{4} \]
(This is also valid since \( 2\pi < \frac{25\pi}{4} < 4\pi \))

3. For \( n = 7 \):
\[ \theta = \frac{\pi}{4} + 7\pi = \frac{\pi}{4} + \frac{28\pi}{4} = \frac{29\pi}{4} \]
(This exceeds \( 4\pi \), so we stop here)

### Step 5: Sum the Valid Solutions
Now, we sum the valid solutions:
\[ \frac{21\pi}{4} + \frac{25\pi}{4} = \frac{46\pi}{4} = \frac{23\pi}{2} \]

### Final Answer
Thus, the sum of all the \( \theta \) that satisfy the equation is:
\[ \frac{23\pi}{2} \]