ANSWER
[tex]{ \sec}(e) = \pm \: \sqrt{\frac{11}{8} } [/tex]
EXPLANATION
We use the Pythagorean Identity,
[tex] { \sec}^{2} (e) = 1 + { \tan}^{2} (e)[/tex]
It was given that,
[tex] { \tan}^{2} (e) = \frac{3}{8} [/tex]
We substitute the values into the identity to obtain,
[tex] { \sec}^{2} (e) = 1 + \frac{3}{8} [/tex]
[tex]{ \sec}^{2} (e) = \frac{11}{8} [/tex]
We take square root of both sides to get,
[tex]{ \sec}(e) = \pm \: \sqrt{\frac{11}{8} } [/tex]
