Question

If $\sec \theta =x+\frac{1}{4x}$, then the value of $\sec \theta +\tan \theta$ is equal to

Answer 1

${\tan }^2\theta ={\sec }^2\theta -1$
$={(x+\frac{1}{4x})}^2-1$
$={(x+\frac{1}{4x})}^2$
$\therefore \tan \theta =\pm (x-\frac{1}{4x})$
$=(x-\frac{1}{4x})or(\frac{1}{4x}-x)$
$\therefore \sec \theta +\tan \theta$ is either 2x or $\frac{1}{2x}$