If sec theta is equal to …

By the given condition of question 
$\sec \theta = x + \frac { 1 } { 4 x }$
$\therefore \quad \tan ^ { 2 } \theta = \sec ^ { 2 } \theta - 1$
$\Rightarrow \quad \tan ^ { 2 } \theta = \left( x + \frac { 1 } { 4 x } \right) ^ { 2 } - 1 = x ^ { 2 } + \frac { 1 } { 16 x ^ { 2 } } + \frac { 1 } { 2 } - 1 = x ^ { 2 } + \frac { 1 } { 16 x ^ { 2 } } - \frac { 1 } { 2 } = \left( x - \frac { 1 } { 4 x } \right) ^ { 2 }$
$\Rightarrow \quad \tan \theta = \pm \left( x - \frac { 1 } { 4 x } \right)$
$\Rightarrow \quad \tan \theta = \left( x - \frac { 1 } { 4 x } \right) \text { or, } \tan \theta = - \left( x - \frac { 1 } { 4 x } \right)$
CASE 1: When $\tan \theta = - \left( x - \frac { 1 } { 4 x } \right) :$ In this case,
$\sec \theta + \tan \theta = x + \frac { 1 } { 4 x } + x - \frac { 1 } { 4 x } = 2 x$
CASE 2: When $\theta = - \left( x - \frac { 1 } { 4 x } \right) :$ In this case,
$\sec \theta + \tan \theta = \left( x + \frac { 1 } { 4 x } \right) - \left( x - \frac { 1 } { 4 x } \right) = \frac { 2 } { 4 x } = \frac { 1 } { 2 x }$
Hence, $\sec \theta + \tan \theta = 2 x \text { or } , \frac { 1 } { 2 x }$