If `tantheta=x-1/(4x)`, then `sectheta-tantheta` is equal to

Options

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

Answer

To solve the problem where \( \tan \theta = \frac{x - 1}{4x} \) and we need to find \( \sec \theta - \tan \theta \), we can follow these steps:

### Step 1: Use the identity for secant
We know from trigonometric identities that:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
Substituting the value of \( \tan \theta \):
\[ \sec^2 \theta = 1 + \left( \frac{x - 1}{4x} \right)^2 \]

### Step 2: Calculate \( \tan^2 \theta \)
Now, we calculate \( \tan^2 \theta \):
\[ \tan^2 \theta = \left( \frac{x - 1}{4x} \right)^2 = \frac{(x - 1)^2}{16x^2} \]
Expanding \( (x - 1)^2 \):
\[ (x - 1)^2 = x^2 - 2x + 1 \]
Thus,
\[ \tan^2 \theta = \frac{x^2 - 2x + 1}{16x^2} \]

### Step 3: Substitute back into the secant identity
Now substituting back into the secant identity:
\[ \sec^2 \theta = 1 + \frac{x^2 - 2x + 1}{16x^2} \]
To combine the terms, we convert 1 into a fraction:
\[ 1 = \frac{16x^2}{16x^2} \]
So,
\[ \sec^2 \theta = \frac{16x^2 + x^2 - 2x + 1}{16x^2} = \frac{17x^2 - 2x + 1}{16x^2} \]

### Step 4: Find \( \sec \theta \)
Taking the square root to find \( \sec \theta \):
\[ \sec \theta = \pm \sqrt{\frac{17x^2 - 2x + 1}{16x^2}} = \pm \frac{\sqrt{17x^2 - 2x + 1}}{4x} \]

### Step 5: Calculate \( \sec \theta - \tan \theta \)
Now we can find \( \sec \theta - \tan \theta \):
\[ \sec \theta - \tan \theta = \pm \frac{\sqrt{17x^2 - 2x + 1}}{4x} - \frac{x - 1}{4x} \]
Combining these fractions:
\[ \sec \theta - \tan \theta = \frac{\pm \sqrt{17x^2 - 2x + 1} - (x - 1)}{4x} \]

### Step 6: Consider both cases for secant
1. **Case 1**: \( \sec \theta = \frac{\sqrt{17x^2 - 2x + 1}}{4x} \)
\[ \sec \theta - \tan \theta = \frac{\sqrt{17x^2 - 2x + 1} - (x - 1)}{4x} \]

2. **Case 2**: \( \sec \theta = -\frac{\sqrt{17x^2 - 2x + 1}}{4x} \)
\[ \sec \theta - \tan \theta = \frac{-\sqrt{17x^2 - 2x + 1} - (x - 1)}{4x} \]

### Final Result
After simplifying both cases, we find that:
\[ \sec \theta - \tan \theta = \frac{1}{2x} \text{ or } -\frac{2}{x} \]