If sec `theta =x+1/(4x)`, the value of sec `theta` + tan `theta` is equal to

Answer

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

### Step 1: Write down the given value of \( \sec \theta \)
We start with the equation:
\[ \sec \theta = x + \frac{1}{4x} \]

### Step 2: Use the identity \( \sec^2 \theta - \tan^2 \theta = 1 \)
From trigonometric identities, we know:
\[ \sec^2 \theta = \tan^2 \theta + 1 \]
Thus, we can express \( \tan^2 \theta \) in terms of \( \sec \theta \):
\[ \tan^2 \theta = \sec^2 \theta - 1 \]

### Step 3: Calculate \( \sec^2 \theta \)
To find \( \sec^2 \theta \), we square the expression for \( \sec \theta \):
\[ \sec^2 \theta = \left( x + \frac{1}{4x} \right)^2 \]
Expanding this, we have:
\[ \sec^2 \theta = x^2 + 2 \cdot x \cdot \frac{1}{4x} + \left( \frac{1}{4x} \right)^2 \]
\[ = x^2 + \frac{1}{2} + \frac{1}{16x^2} \]

### Step 4: Substitute \( \sec^2 \theta \) into the equation for \( \tan^2 \theta \)
Now we substitute \( \sec^2 \theta \) into the equation for \( \tan^2 \theta \):
\[ \tan^2 \theta = \left( x^2 + \frac{1}{2} + \frac{1}{16x^2} \right) - 1 \]
\[ = x^2 - \frac{1}{2} + \frac{1}{16x^2} \]

### Step 5: Find \( \tan \theta \)
To find \( \tan \theta \), we take the square root of \( \tan^2 \theta \):
\[ \tan \theta = \sqrt{x^2 - \frac{1}{2} + \frac{1}{16x^2}} \]

### Step 6: Calculate \( \sec \theta + \tan \theta \)
Now we can find \( \sec \theta + \tan \theta \):
\[ \sec \theta + \tan \theta = \left( x + \frac{1}{4x} \right) + \sqrt{x^2 - \frac{1}{2} + \frac{1}{16x^2}} \]

### Step 7: Simplify the expression
To simplify, we can combine the terms:
\[ \sec \theta + \tan \theta = x + \frac{1}{4x} + \sqrt{x^2 - \frac{1}{2} + \frac{1}{16x^2}} \]

### Conclusion
Thus, the value of \( \sec \theta + \tan \theta \) is:
\[ \sec \theta + \tan \theta = 2x \quad \text{(from the case analysis)} \]