Answer:
[tex] \sec \theta = \pm \frac{ \sqrt{ {x}^{2} + 16} }{4} [/tex]
Step-by-step explanation:
[tex] \because \: x = 4 \tan \theta \\ \therefore\frac{x}{4} = \tan \theta....(1) \\ \\ \because \: { \sec}^{2} \theta = 1 + { \tan}^{2} \theta \\ \therefore \: \sec \theta = \pm\sqrt{1 + { \tan}^{2} \theta } \\ \therefore \: \sec \theta = \pm\sqrt{1 + { \bigg( \frac{x}{4} \bigg)}^{2} } \\ \therefore \: \sec \theta = \pm\sqrt{1 + { \frac{x^{2}}{16} } } \\ \therefore \: \sec \theta = \pm\sqrt{{ \frac{16 + x^{2}}{16} } } \\ \therefore \: \sec \theta = \pm \frac{ \sqrt{ {x}^{2} + 16} }{4} [/tex]
