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[tex]\sec^2\theta - \tan^2 \theta = 1[/tex]
[tex]\therefore \tan\theta = \sqrt{\sec^2 \theta - 1}[/tex]
[tex]\displaystyle \therefore \tan\theta = \sqrt{\left(x + \frac{1}{4x}\right)^2 - 1}[/tex]
[tex]\displaystyle = \sqrt{x^2 + \frac{1}{16x^2} + \frac{1}{2} - 1}[/tex]
[tex]\displaystyle = \sqrt{x^2 + \frac{1}{16x^2} - \frac{1}{2}}[/tex]
[tex]\displaystyle = \sqrt{\left(x - \frac{1}{4x}\right)^2}[/tex]
[tex]\displaystyle = \pm \left(x- \frac{1}{4x}\right)[/tex]
Hence,
[tex]\displaystyle \sec\theta + \tan\theta = 2x \text{ or } \frac{1}{2x}[/tex]