Question

If \( \sec \theta = a + \frac{1}{4a}, 0^\circ<\theta<90^\circ \), then \( \csc \theta + \cot \theta = \):

• A. \( \frac{2a}{2a + 1} \)
• B. \( \frac{4a}{2a - 1} \)
• C. \( \frac{2a + 1}{2a - 1} \)
• D. \( \frac{4a - 1}{2a + 1} \)

Hint:

For expressions involving trigonometric identities, try to simplify using known identities, such as \( \sec^2 \theta - \tan^2 \theta = 1 \), and solve for the required terms.

Answer 1

The Correct Option is C

We are given that \( \sec \theta = a + \frac{1}{4a} \). Our goal is to find \( \csc \theta + \cot \theta \). ### Step 1: Use the identity \( \sec^2 \theta - \tan^2 \theta = 1 \). We start with the identity: \[ \sec^2 \theta - \tan^2 \theta = 1 \] Using the given equation for \( \sec \theta \): \[ \sec \theta = a + \frac{1}{4a} \] Square both sides: \[ \sec^2 \theta = \left( a + \frac{1}{4a} \right)^2 \] \[ = a^2 + 2 \cdot a \cdot \frac{1}{4a} + \left( \frac{1}{4a} \right)^2 \] \[ = a^2 + \frac{1}{2} + \frac{1}{16a^2} \] Thus, \( \sec^2 \theta = a^2 + \frac{1}{2} + \frac{1}{16a^2} \). ### Step 2: Use the identity for \( \csc^2 \theta \). Next, use the identity \( \csc^2 \theta = 1 + \cot^2 \theta \), and write \( \csc \theta \) and \( \cot \theta \) in terms of \( \sec \theta \). After simplifying the equations, we find: \[ \csc \theta + \cot \theta = \frac{2a + 1}{2a - 1} \] Thus, the correct answer is \( \frac{2a + 1}{2a - 1} \).