Question - Proving Trigonometric Identity Involving Secant and Tangent Functions

Given: \(\sec(\theta) = x + \frac{1}{4x}\)

To prove: \(\sec^2(\theta) + \tan^2(\theta) = 2x \text{ or } \frac{1}{2x}\)

We know that: \(\sec^2(\theta) = 1 + \tan^2(\theta)\)

So, \(\sec^2(\theta) + \tan^2(\theta) = 2 \sec^2(\theta) - 1\)

Using given \(\sec(\theta) = x + \frac{1}{4x}\), we get:

\(\sec^2(\theta) = \left(x + \frac{1}{4x}\right)^2 = x^2 + 2 \cdot x \cdot \frac{1}{4x} + \frac{1}{16x^2}\)

\(\sec^2(\theta) = x^2 + \frac{1}{2} + \frac{1}{16x^2}\)

Hence \(\sec^2(\theta) + \tan^2(\theta) = 2x^2 + 1 + \frac{1}{8x^2} - 1\)

\(\sec^2(\theta) + \tan^2(\theta) = 2x^2 + \frac{1}{8x^2}\)

We realize there is a mistake because we cannot get \(2x\) or \(\frac{1}{2x}\) directly from \(2x^2 + \frac{1}{8x^2}\), thus the original statement seems incorrect. We need to reassess the problem or check the given identity.