Respuesta :
Answer:
B) Plus-or-minus StartRoot eleven-eighths EndFraction
[tex] { \tan }^{2} \theta = \frac{3}{8} \\ but : { \tan }^{2} \theta = { \sec }^{2} \theta - 1 \\ \therefore { \sec}^{2} \theta - 1 = \frac{3}{8} \\ { \sec }^{2} \theta = \frac{11}{8} \\ \sec( \theta) = \sqrt{ \frac{11}{8} } [/tex]
Answer:
B) Plus-or-minus StartRoot eleven-eighths EndFraction
Step-by-step explanation:
