Answer
$\sin\theta=\frac{y}{r}=\frac{\sqrt {15}}{4}$
$\cos\theta=\frac{x}{r}=\frac{-1}{4}$
$\tan\theta=\frac{y}{x}=\frac{\sqrt {15} }{-1}=-\sqrt {15}$
$\cot\theta=\frac{x}{y}=\frac{-1}{\sqrt {15}}=\frac{-\sqrt {15}}{15}$
$\sec\theta=\frac{r}{x}=\frac{4}{-1}=-4$
$\csc\theta=\frac{r}{y}=\frac{4}{\sqrt {15}}=\frac{4\sqrt {15}}{15}$
Work Step by Step
1. II quadrant : x negative and y is positive, because sec is negative and sin is positive.
x=-1 and r=4
2. Calculate y using distance formula
$4=\sqrt {(y)^{2}+(-1)^{2}}$
$16=1+y^{2}$
$y=\sqrt {15}$
3. Insert the values to find trig functions
$\sin\theta=\frac{y}{r}=\frac{\sqrt {15}}{4}$
$\cos\theta=\frac{x}{r}=\frac{-1}{4}$
$\tan\theta=\frac{y}{x}=\frac{\sqrt {15} }{-1}=-\sqrt {15}$
$\cot\theta=\frac{x}{y}=\frac{-1}{\sqrt {15}}=\frac{-\sqrt {15}}{15}$
$\sec\theta=\frac{r}{x}=\frac{4}{-1}=-4$
$\csc\theta=\frac{r}{y}=\frac{4}{\sqrt {15}}=\frac{4\sqrt {15}}{15}$
