\(\cos{A}= \frac{1}{2} \Rightarrow A=\pm \frac{\pi}{3} +2k\pi,k \in Z\)
A ∈ IV quadrant ⇒ \(A=- \frac{\pi}{3}\)
\(\sec{ \frac{A}{2}}= \frac{1}{\cos{\frac{A}{2}}}= \frac{1}{\cos{\frac{- \frac{\pi}{3} }{2}}}= \frac{1}{\cos{(- \frac{\pi}{6} )}}= \frac{1}{\cos{ \frac{\pi}{6}}}= \frac{1}{ \frac{ \sqrt{3} }{2} } = \frac{2}{ \sqrt{3} }\)