If `(1)/( cos theta) = a + (1)/(4a)` , then the value of `(tan theta + (1)/( cos theta))` is

Options

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

Answer

To solve the equation \( \frac{1}{\cos \theta} = a + \frac{1}{4a} \) and find the value of \( \tan \theta + \frac{1}{\cos \theta} \), we can follow these steps:

### Step 1: Rewrite the equation
We know that \( \sec \theta = \frac{1}{\cos \theta} \). Thus, we can rewrite the given equation as:
\[ \sec \theta = a + \frac{1}{4a} \]

### Step 2: Express \( \tan \theta \) in terms of \( \sec \theta \)
Using the identity \( \tan^2 \theta + 1 = \sec^2 \theta \), we can express \( \tan \theta \):
\[ \tan^2 \theta = \sec^2 \theta - 1 \]
Taking the square root gives:
\[ \tan \theta = \sqrt{\sec^2 \theta - 1} \]

### Step 3: Substitute \( \sec \theta \)
Now substitute \( \sec \theta = a + \frac{1}{4a} \) into the equation:
\[ \tan \theta = \sqrt{\left(a + \frac{1}{4a}\right)^2 - 1} \]

### Step 4: Simplify \( \tan \theta \)
Now, we need to simplify \( \left(a + \frac{1}{4a}\right)^2 - 1 \):
\[ \left(a + \frac{1}{4a}\right)^2 = a^2 + 2 \cdot a \cdot \frac{1}{4a} + \left(\frac{1}{4a}\right)^2 = a^2 + \frac{1}{2} + \frac{1}{16a^2} \]
Thus,
\[ \tan \theta = \sqrt{a^2 + \frac{1}{2} + \frac{1}{16a^2} - 1} = \sqrt{a^2 - \frac{1}{2} + \frac{1}{16a^2}} \]

### Step 5: Find \( \tan \theta + \sec \theta \)
Now, we can find \( \tan \theta + \sec \theta \):
\[ \tan \theta + \sec \theta = \sqrt{a^2 - \frac{1}{2} + \frac{1}{16a^2}} + \left(a + \frac{1}{4a}\right) \]

### Step 6: Combine the terms
Combine the terms:
\[ \tan \theta + \sec \theta = \sqrt{a^2 - \frac{1}{2} + \frac{1}{16a^2}} + a + \frac{1}{4a} \]

### Final Step: Conclusion
After simplifying, we find that:
\[ \tan \theta + \sec \theta = 2a \]

Thus, the value of \( \tan \theta + \frac{1}{\cos \theta} \) is \( 2a \).

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