Given secant of theta is equal to the square root of 6 over 3 comma what is cos?

Given that
[tex]\sec \text{ }\theta\text{ = }\frac{\sqrt[]{6}}{3}[/tex]Required: cos
From the reciprocal of trigonometric function,
[tex]\sec \text{ }\theta\text{ = }\frac{1}{\cos \text{ }\theta}[/tex]Thus, we have
[tex]\frac{1}{\cos \text{ }\theta}\text{ = }\frac{\sqrt[]{6}}{3}[/tex]Cross-multiply, we have
[tex]\begin{gathered} \sqrt[]{6}\text{ }\times\text{ cos }\theta\text{ = 3}\times1 \\ \sqrt[]{6}\text{ cos }\theta\text{ =3} \end{gathered}[/tex]Divide both sides by the coefficient of cos θ.
[tex]\begin{gathered} \frac{\sqrt[]{6}\text{ cos }\theta}{\sqrt[]{6}}\text{ =}\frac{\text{3}}{\sqrt[]{6}} \\ \Rightarrow\cos \text{ }\theta\text{ = }\frac{\text{3}}{\sqrt[]{6}} \\ \end{gathered}[/tex]Rationalizing the resulting surd, we have
[tex]\begin{gathered} \text{ }\frac{3}{\sqrt[]{6}}\times\frac{\sqrt[]{6}}{\sqrt[]{6}} \\ =\frac{3\times\sqrt[]{6}}{\sqrt[]{6}\times\sqrt[]{6}}=\frac{3\sqrt[]{6}}{6} \\ =\frac{\sqrt[]{6}}{2} \\ \text{Thus,} \\ \cos \text{ }\theta\text{ = }\frac{\sqrt[]{6}}{2} \end{gathered}[/tex]Hence, cos θ is evaluated to be
[tex]\frac{\sqrt[]{6}}{2}[/tex]The second option is the correct answer.