## Precalculus (6th Edition) Blitzer

The trigonometric functions are: $\sin \theta =-\frac{\sqrt{10}}{10}\text{; cos}\theta \text{=}-\frac{3\sqrt{10}}{10}\text{; tan}\theta \text{=}\frac{1}{3};\text{ csc}\theta \text{=}-\sqrt{10};\text{ sec}\theta \text{=}-\frac{\sqrt{10}}{3}$.
We know that $\cot \theta >0\text{ and cos}\theta <0$ , so $\theta$ lies in quadrant III. So, in quadrant III, x and y are both negative. Here, \begin{align} & \cot \theta =\frac{3}{1} \\ & =\frac{x}{y} \\ & =\frac{-3}{-1} \end{align} Therefore, $x=-3\text{ and }y=-1$. Now, calculate r as: \begin{align} & {{r}^{2}}={{x}^{2}}+{{y}^{2}} \\ & r=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -1 \right)}^{2}}} \\ & r=\sqrt{9+1} \\ & r=\sqrt{10} \end{align} Now, evaluate the sine function as follows: \begin{align} & \sin \theta =\frac{y}{r} \\ & =\frac{-1}{\sqrt{10}} \\ & =\frac{-1\cdot \sqrt{10}}{\sqrt{10}\cdot \sqrt{10}} \\ & =-\frac{\sqrt{10}}{10} \end{align} Then, evaluate the cosine function as follows: \begin{align} & \cos \theta =\frac{x}{r} \\ & =\frac{-3}{\sqrt{10}} \\ & =-\frac{3\cdot \sqrt{10}}{\sqrt{10}\cdot \sqrt{10}} \\ & =-\frac{3\sqrt{10}}{10} \end{align} Then, evaluate the tangent function as follows: \begin{align} & \tan \theta =\frac{y}{x} \\ & =\frac{-1}{-3} \\ & =\frac{1}{3} \end{align} Then, evaluate the cosecant function as follows: \begin{align} & \csc \theta =\frac{r}{y} \\ & =\frac{\sqrt{10}}{-1} \\ & =-\sqrt{10} \end{align} Finally evaluate the secant function as follows: \begin{align} & \sec \theta =\frac{r}{x} \\ & =\frac{\sqrt{10}}{-3} \\ & =-\frac{\sqrt{10}}{3} \end{align}