We need to find cos of theta using the equation of secant of theta:
Now, we need to use the next trigonometric function:
[tex]\begin{gathered} \sec \theta=\frac{1}{\cos\theta} \\ \text{Then, we can replace} \\ \frac{1}{\cos\theta}=\frac{\sqrt[]{6}}{3} \\ \text{Solve for cos}\theta\colon \\ \text{Multiply both sides by 3:} \\ 3(\frac{1}{\cos\theta})=3(\frac{\sqrt[]{6}}{3}) \\ \frac{3}{\text{cos}\theta}=\sqrt[]{6} \\ \text{Multiply both sides by cos}\theta\colon \\ \cos \theta\cdot\frac{3}{\cos\theta}=\cos \theta\cdot\sqrt[]{6} \\ 3=\cos \theta\cdot\sqrt[]{6} \\ \text{Divide both sides by }\sqrt[]{6} \\ \frac{3}{\sqrt[]{6}}=\frac{\cos\theta\cdot\sqrt[]{6}}{\sqrt[]{6}} \\ \frac{3}{\sqrt[]{6}}=\cos \theta \\ it\text{ also can be expressed as :} \\ \cos \theta=\frac{\sqrt[]{6}}{2} \end{gathered}[/tex]
Therefore, the correct answer is the second one.
[tex]\begin{gathered} To\text{ get the answer we need to cancel the square root on the denominator:} \\ \frac{3}{\sqrt[]{6}}\cdot\frac{\sqrt[]{6}}{\sqrt[]{6}}=\frac{3\sqrt[]{6}}{(\sqrt[]{6})^2}=\frac{3\sqrt[]{6}}{6} \\ \text{Simplify by 3:} \\ \frac{3\sqrt[]{6}}{6}=\frac{1\sqrt[]{6}}{2}=\frac{\sqrt[]{6}}{2} \end{gathered}[/tex]
