If ` cosec theta =x + 1/(4x) ` then the value of `cosec theta + cot theta ` is

Options

A. 2x
B. `-2x`
C. `2/x`
D. `-1/(2x)`

Explanation

To solve the problem, we need to find the value of \( \csc \theta + \cot \theta \) given that \( \csc \theta = x + \frac{1}{4x} \).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ \csc \theta = x + \frac{1}{4x} \]

2. **Recall the relationship between cosecant and cotangent**:
We know that:
\[ \csc^2 \theta = 1 + \cot^2 \theta \]
Therefore, we can express \( \cot^2 \theta \) in terms of \( \csc \theta \):
\[ \cot^2 \theta = \csc^2 \theta - 1 \]

3. **Calculate \( \csc^2 \theta \)**:
First, we need to square \( \csc \theta \):
\[ \csc^2 \theta = \left(x + \frac{1}{4x}\right)^2 \]
Expanding this:
\[ \csc^2 \theta = x^2 + 2 \cdot x \cdot \frac{1}{4x} + \left(\frac{1}{4x}\right)^2 \]
\[ = x^2 + \frac{1}{2} + \frac{1}{16x^2} \]

4. **Substitute into the cotangent equation**:
Now, substitute \( \csc^2 \theta \) into the cotangent equation:
\[ \cot^2 \theta = \left(x^2 + \frac{1}{2} + \frac{1}{16x^2}\right) - 1 \]
\[ = x^2 - \frac{1}{2} + \frac{1}{16x^2} \]

5. **Find \( \cot \theta \)**:
Taking the square root:
\[ \cot \theta = \sqrt{x^2 - \frac{1}{2} + \frac{1}{16x^2}} \]

6. **Calculate \( \csc \theta + \cot \theta \)**:
Now we can find \( \csc \theta + \cot \theta \):
\[ \csc \theta + \cot \theta = \left(x + \frac{1}{4x}\right) + \sqrt{x^2 - \frac{1}{2} + \frac{1}{16x^2}} \]

7. **Simplify the expression**:
However, we notice that we can simplify this further. We have:
\[ \cot \theta = x - \frac{1}{4x} \]
Therefore:
\[ \csc \theta + \cot \theta = \left(x + \frac{1}{4x}\right) + \left(x - \frac{1}{4x}\right) \]
\[ = 2x \]

### Final Answer:
Thus, the value of \( \csc \theta + \cot \theta \) is:
\[ \boxed{2x} \]