If `tan theta =a - ( 1)/( 4a) `, then `sec theta - tan theta=`

Options

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

Answer

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

### Step 1: Find \( \sec^2 \theta \)
Using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \), we can express \( \sec^2 \theta \) in terms of \( \tan \theta \).

\[ \tan \theta = a - \frac{1}{4a} \]

Now, we square \( \tan \theta \):

\[ \tan^2 \theta = \left(a - \frac{1}{4a}\right)^2 = a^2 - 2 \cdot a \cdot \frac{1}{4a} + \left(\frac{1}{4a}\right)^2 \]

Calculating this gives:

\[ \tan^2 \theta = a^2 - \frac{1}{2} + \frac{1}{16a^2} \]

Now, substituting this into the identity for \( \sec^2 \theta \):

\[ \sec^2 \theta = 1 + \left(a^2 - \frac{1}{2} + \frac{1}{16a^2}\right) \]

Simplifying this:

\[ \sec^2 \theta = 1 + a^2 - \frac{1}{2} + \frac{1}{16a^2} = a^2 + \frac{1}{2} + \frac{1}{16a^2} \]

### Step 2: Find \( \sec \theta \)
To find \( \sec \theta \), we take the square root of \( \sec^2 \theta \):

\[ \sec \theta = \sqrt{a^2 + \frac{1}{2} + \frac{1}{16a^2}} \]

This can be rewritten as:

\[ \sec \theta = \sqrt{\left(a + \frac{1}{4a}\right)^2} \]

Thus, we have:

\[ \sec \theta = a + \frac{1}{4a} \quad \text{(taking the positive root)} \]

### Step 3: Calculate \( \sec \theta - \tan \theta \)
Now we can find \( \sec \theta - \tan \theta \):

\[ \sec \theta - \tan \theta = \left(a + \frac{1}{4a}\right) - \left(a - \frac{1}{4a}\right) \]

Simplifying this gives:

\[ \sec \theta - \tan \theta = a + \frac{1}{4a} - a + \frac{1}{4a} = \frac{1}{2a} \]

### Step 4: Consider the negative root
If we consider the negative root for \( \sec \theta \):

\[ \sec \theta = -\left(a + \frac{1}{4a}\right) \]

Then:

\[ \sec \theta - \tan \theta = -\left(a + \frac{1}{4a}\right) - \left(a - \frac{1}{4a}\right) \]

This simplifies to:

\[ \sec \theta - \tan \theta = -2a \]

### Final Result
Thus, the two possible values for \( \sec \theta - \tan \theta \) are:

\[ \sec \theta - \tan \theta = \frac{1}{2a} \quad \text{or} \quad -2a \]