Question

If \( \sec \theta = a + \frac{1}{4a^2} \), \( 0^\circ<\theta<90^\circ \), 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:

Try rationalizing given expressions and use the identity \( \csc \theta + \cot \theta = \frac{1 + \cos \theta}{\sin \theta} \). It helps convert complex expressions into a solvable form.

Answer 1

The Correct Option is C

Given: \[ \sec \theta = a + \frac{1}{4a^2} \Rightarrow \cos \theta = \frac{1}{a + \frac{1}{4a^2}} = \frac{4a^2}{4a^3 + 1} \] Using identity: \[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left( \frac{4a^2}{4a^3 + 1} \right)^2 \Rightarrow \sin \theta = \sqrt{1 - \left( \frac{16a^4}{(4a^3 + 1)^2} \right)} = \frac{\sqrt{(4a^3 + 1)^2 - 16a^4}}{4a^3 + 1} \] Let’s define \(\csc \theta = \frac{1}{\sin \theta}\), and \(\cot \theta = \frac{\cos \theta}{\sin \theta}\), so: \[ \csc \theta + \cot \theta = \frac{1 + \cos \theta}{\sin \theta} \Rightarrow \frac{1 + \frac{4a^2}{4a^3 + 1}}{\frac{\sqrt{(4a^3 + 1)^2 - 16a^4}}{4a^3 + 1}} = \frac{4a^3 + 1 + 4a^2}{\sqrt{(4a^3 + 1)^2 - 16a^4}} \] Eventually, this simplifies to \( \frac{2a + 1}{2a - 1} \)