# 005 Sample Final A, Question 12

Question Given that ${\displaystyle \sec(\theta )=-2}$ and ${\displaystyle \tan(\theta )>0}$, find the exact values of the remaining trig functions.

Foundations
1) Which quadrant is ${\displaystyle \theta }$ in?
2) Which trig functions are positive in this quadrant?
3) What are the side lengths of the triangle associated to ${\displaystyle \theta ?}$
1) ${\displaystyle \theta }$ is in the third quadrant. We know it is in the second or third quadrant since ${\displaystyle \cos }$ is negative. Since \${\displaystyle \tan }$ is positive ${\displaystyle \theta }$ is in the third quadrant.
2) ${\displaystyle \tan }$ and ${\displaystyle \cot }$ are both positive in this quadrant. All other trig functions are negative.
3) The side lengths are 2, 1, and ${\displaystyle {\sqrt {3}}.}$
Step 1:
Since ${\displaystyle \sec(\theta )=-2}$, we have ${\displaystyle \cos(\theta )={\frac {1}{\sec(\theta )}}={\frac {-1}{2}}}$.
Step 2:
We look for solutions to ${\displaystyle \theta }$ on the unit circle. The two angles on the unit circle with ${\displaystyle \cos(\theta )={\frac {-1}{2}}}$ are ${\displaystyle \theta ={\frac {2\pi }{3}}}$ and ${\displaystyle \theta ={\frac {4\pi }{3}}}$.
But, ${\displaystyle \tan \left({\frac {2\pi }{3}}\right)=-{\sqrt {3}}}$. Since ${\displaystyle \tan(\theta )>0}$. we must have ${\displaystyle \theta ={\frac {4\pi }{3}}}$.
Step 3:
The remaining values of the trig functions are
${\displaystyle \sin(\theta )=\sin \left({\frac {4\pi }{3}}\right)={\frac {-{\sqrt {3}}}{2}}}$,
${\displaystyle \tan(\theta )=\tan \left({\frac {4\pi }{3}}\right)={\sqrt {3}}}$
${\displaystyle \csc(\theta )=\csc \left({\frac {4\pi }{3}}\right)={\frac {-2{\sqrt {3}}}{3}}}$ and
${\displaystyle \cot(\theta )=\cot \left({\frac {4\pi }{3}}\right)={\frac {\sqrt {3}}{3}}}$

${\displaystyle \sin(\theta )=={\frac {-{\sqrt {3}}}{2}}}$
${\displaystyle \cos(\theta )={\frac {-1}{2}}}$
${\displaystyle \tan(\theta )={\sqrt {3}}}$
${\displaystyle \csc(\theta )={\frac {-2{\sqrt {3}}}{3}}}$
${\displaystyle \cot(\theta )={\frac {\sqrt {3}}{3}}}$