If tanA = x – 1/4x, then what will be the value of secA – tanA?

Options

A. 2x
B. x
C. 1/2x
D. `1//2x^2`

Answer

To solve the problem, we need to find the value of \( \sec A - \tan A \) given that \( \tan A = x - \frac{1}{4x} \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \tan A = x - \frac{1}{4x} \]

2. **Use the identity for \( \sec^2 A \):**
We know that:
\[ \sec^2 A = 1 + \tan^2 A \]

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

4. **Substitute \( \tan^2 A \) into the \( \sec^2 A \) formula:**
\[ \sec^2 A = 1 + \left(x^2 - \frac{1}{2} + \frac{1}{16x^2}\right) \]
Simplifying this gives:
\[ \sec^2 A = x^2 + \frac{1}{2} + \frac{1}{16x^2} \]

5. **Now, calculate \( \sec A \):**
Since \( \sec A = \sqrt{\sec^2 A} \), we have:
\[ \sec A = \sqrt{x^2 + \frac{1}{2} + \frac{1}{16x^2}} \]

6. **Now, find \( \sec A - \tan A \):**
We need to compute:
\[ \sec A - \tan A = \sqrt{x^2 + \frac{1}{2} + \frac{1}{16x^2}} - \left(x - \frac{1}{4x}\right) \]
This simplifies to:
\[ \sec A - \tan A = \sqrt{x^2 + \frac{1}{2} + \frac{1}{16x^2}} - x + \frac{1}{4x} \]

7. **Combine the terms:**
To combine these terms, we can express \( \sec A - \tan A \) in a single fraction. However, we can also evaluate it directly based on the earlier steps.

8. **Final simplification:**
After simplification, we find that:
\[ \sec A - \tan A = \frac{1}{2x} \]

### Conclusion:
Thus, the value of \( \sec A - \tan A \) is:
\[ \frac{1}{2x} \]