check the picture below, those are the x,y pairs or cosine, sine values pair.
[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad
cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\
-------------------------------\\\\
cos\left( \frac{7\pi }{6} \right)=\cfrac{\stackrel{adjacent}{-\sqrt{3}}}{\stackrel{hypotenuse}{2}}\qquad \qquad sin\left( \frac{7\pi }{6} \right)=\cfrac{\stackrel{opposite}{1}}{\stackrel{hypotenuse}{2}}[/tex]
now, this is from the Unit Circle, and therefore the hypotenuse or radius wil be 1.
[tex]\bf \begin{cases}
adjacent=&-\frac{\sqrt{3}}{2}\\
opposite=&\frac{1}{2}\\
hypotenuse=&1
\end{cases}\\\\
-------------------------------\\\\
sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\quad
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\\\\\\
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\qquad
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\quad
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}[/tex]
so, just plug in those values... hmmm lemme do the tangent, so you see the division of two fractions.
[tex]\bf tan\left( \frac{7\pi }{6} \right)=\cfrac{sin\left( \frac{7\pi }{6} \right)}{cos\left( \frac{7\pi }{6} \right)}\implies tan\left( \frac{7\pi }{6} \right)=\cfrac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}\implies tan\left( \frac{7\pi }{6} \right)=\cfrac{1}{2}\cdot \cfrac{2}{-\sqrt{3}}[/tex]
[tex]\bf tan\left( \frac{7\pi }{6} \right)=\cfrac{1}{-\sqrt{3}}\impliedby \textit{now, let's \underline{rationalize} the denominator}
\\\\\\
\cfrac{1}{-\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{\sqrt{3}}{-(\sqrt{3})^2}\implies -\cfrac{\sqrt{3}}{3}[/tex]
