Answer
$\sin\theta=\frac{y}{r}=\frac{1}{2}$
$\cos\theta=\frac{x}{r}=\frac{-\sqrt {3}}{2}$
$\tan\theta=\frac{y}{x}=\frac{1 }{-\sqrt {3}}=-\frac{\sqrt 3}{3}$
$\cot\theta=\frac{x}{y}=\frac{-\sqrt {3}}{1}=-\sqrt 3$
$\sec\theta=\frac{r}{x}=\frac{2}{-\sqrt {3}}=-\frac{2\sqrt 3}{3}$
$\csc\theta=\frac{r}{y}=\frac{2}{1}=2$
Work Step by Step
1.Quadrant II --> x is negative and y is positive
y=1 and r=2
2. FInd x by using distance formula
$2=\sqrt {(x)^{2}+(1)^{2}}$
$4=1+x^{2}$
$x=-\sqrt 3$
3. Plug the values to find trig functions
$\sin\theta=\frac{y}{r}=\frac{1}{2}$
$\cos\theta=\frac{x}{r}=\frac{-\sqrt {3}}{2}$
$\tan\theta=\frac{y}{x}=\frac{1 }{-\sqrt {3}}=-\frac{\sqrt 3}{3}$
$\cot\theta=\frac{x}{y}=\frac{-\sqrt {3}}{1}=-\sqrt 3$
$\sec\theta=\frac{r}{x}=\frac{2}{-\sqrt {3}}=-\frac{2\sqrt 3}{3}$
$\csc\theta=\frac{r}{y}=\frac{2}{1}=2$
