Trigonometry (10th Edition)

$\cot\theta=\sqrt 3$
1.We get this information from the given cosecant value $y=-1$ $r=2$ 2. We can use y and r to find x (we need to remember that sign of x depends on quadrant, since it ends in quadrant III, x must be negative) $r=\sqrt {y^{2}+x^{2}}$ $x=-\sqrt {r^{2}-y^{2}}=-\sqrt {(2)^{2}-(-1)^{2}}=-\sqrt {3}$ 4. Then insert values, $\cot\theta=\sqrt 3$