Question

solve this problem ​

Answer 1

$\sec^2\theta - \tan^2 \theta = 1$

$\therefore \tan\theta = \sqrt{\sec^2 \theta - 1}$

$\displaystyle \therefore \tan\theta = \sqrt{\left(x + \frac{1}{4x}\right)^2 - 1}$

$\displaystyle = \sqrt{x^2 + \frac{1}{16x^2} + \frac{1}{2} - 1}$

$\displaystyle = \sqrt{x^2 + \frac{1}{16x^2} - \frac{1}{2}}$

$\displaystyle = \sqrt{\left(x - \frac{1}{4x}\right)^2}$

$\displaystyle = \pm \left(x- \frac{1}{4x}\right)$

Hence,

$\displaystyle \sec\theta + \tan\theta = 2x \text{ or } \frac{1}{2x}$