Step 1
Draw the unit circle required
Step 2
Find the value sec(7π/6) in cosine
[tex]\begin{gathered} \sec (\frac{7\pi}{6})=\frac{1}{cos(\frac{7\pi}{6})} \\ \sec (x)=\frac{1}{cos(x)} \end{gathered}[/tex]Step 3
Find cos(7π/6)
The trigonometric unit circle and a trigonometric table gives;
[tex]\begin{gathered} \cos (\frac{7\pi}{6})=\cos (\frac{\pi}{6}+\pi) \\ \cos (\frac{7\pi}{6})=\text{cos}(\frac{\pi}{6})\cos (\pi)-\sin (\frac{\pi}{6})sin\pi=-\cos (\frac{\pi}{6}) \\ \cos (\frac{7\pi}{6})=\frac{\sqrt[]{3}}{2}(-1)-(\frac{1}{2})(0)=-\frac{\sqrt[]{3}}{2} \\ \cos (\frac{7\pi}{6})=-\frac{\sqrt[]{3}}{2} \end{gathered}[/tex]Step 4
Find sec(7π/6)
[tex]\begin{gathered} \sec (x)=\frac{1}{cos(x)} \\ \text{sec}(\frac{7\pi}{6})=\frac{1}{\cos (\frac{7\pi}{6})} \\ \text{sec}(\frac{7\pi}{6})=\frac{1}{-\frac{\sqrt[]{3}}{2}} \\ \text{sec}(\frac{7\pi}{6})=-\frac{2}{\sqrt[]{3}} \end{gathered}[/tex]Step 5
Rationalize the denominator
[tex]\begin{gathered} \sec (\frac{7\pi}{6})=-\frac{2}{\sqrt[]{3}}\times\frac{\sqrt[]{3}}{\sqrt[]{3}} \\ \sec (\frac{7\pi}{6})=-\frac{2\sqrt[]{3}}{\sqrt[]{9}} \\ \sec (\frac{7\pi}{6})=-\frac{2\sqrt[]{3}}{3} \end{gathered}[/tex]Hence,
[tex]\sec (\frac{7\pi}{6})=-\frac{2\sqrt[]{3}}{3}[/tex]
