Answer:
[tex] \tan \theta = \boxed {2} \sqrt{\boxed{6}}[/tex]
Step-by-step explanation:
[tex] \sec \theta = 5 \: \: (given) \\ \because \: \tan^{2} \theta = { \sec}^{2} \theta - 1\\ \therefore \: \tan \theta = \pm \sqrt{{ \sec}^{2} \theta - 1} \\ = \pm \sqrt{{5}^{2} - 1} \\ = \pm \sqrt{25 - 1} \\ = \pm \sqrt{24} \\ \therefore \: \tan \theta = \pm 2\sqrt{6} \\ \because \theta \: lies \: in \: the \: first \: quadrant \\ \ \therefore \: \tan \theta = \boxed {2} \sqrt{\boxed{6}} [/tex]
