If `sectheta=a+(1)/(4a)`, and `0`<`theta`<`pi/2` find `sectheta+tantheta`.

Options

A. a
B. 2a
C. 3a
D. 4a

Answer

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

### Step-by-Step Solution:

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

2. **Use the identity \( \sec^2 \theta - \tan^2 \theta = 1 \):**
\[ \tan^2 \theta = \sec^2 \theta - 1 \]

3. **Substitute \( \sec \theta \) into the identity:**
\[ \tan^2 \theta = \left(a + \frac{1}{4a}\right)^2 - 1 \]

4. **Expand \( \left(a + \frac{1}{4a}\right)^2 \):**
\[ \tan^2 \theta = \left(a^2 + 2 \cdot a \cdot \frac{1}{4a} + \frac{1}{16a^2}\right) - 1 \]
\[ = a^2 + \frac{1}{2} + \frac{1}{16a^2} - 1 \]
\[ = a^2 - \frac{1}{2} + \frac{1}{16a^2} \]

5. **Now we have \( \tan^2 \theta \):**
\[ \tan^2 \theta = a^2 - \frac{1}{2} + \frac{1}{16a^2} \]

6. **Now, we need to find \( \sec \theta + \tan \theta \):**
\[ \sec \theta + \tan \theta = \left(a + \frac{1}{4a}\right) + \sqrt{a^2 - \frac{1}{2} + \frac{1}{16a^2}} \]

7. **To simplify \( \tan \theta \), we can express it as:**
\[ \tan \theta = \sqrt{a^2 - \frac{1}{2} + \frac{1}{16a^2}} = \sqrt{\left(a - \frac{1}{4a}\right)^2} \]
\[ = a - \frac{1}{4a} \]

8. **Thus, we can write:**
\[ \sec \theta + \tan \theta = \left(a + \frac{1}{4a}\right) + \left(a - \frac{1}{4a}\right) \]
\[ = 2a \]

### Final Answer:
\[ \sec \theta + \tan \theta = 2a \]