# Question #978dc

Aug 24, 2016

#### Explanation:

Given that$\left(- \sqrt{3} , 1\right)$ is the cartesian coordinate of the point on the terminal side of the angle $\theta$.It is in the third quadrant.

Here $x = - \sqrt{3} \mathmr{and} y = 1$

So lrngth of the terminal side
$r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{\left(- \sqrt{3}\right) + {1}^{2}} = 2$

$\text{Here in respect of } \theta$
$x \to \text{adjacent}$

$y \to \text{opposite}$

$r \to \text{hypotenuse}$

$\sin \theta = \frac{y}{r} = \frac{1}{2}$

$\cos \theta = \frac{x}{r} = - \frac{\sqrt{3}}{2}$

$\tan \theta = \frac{y}{x} = \frac{1}{- \sqrt{3}} - \frac{1}{\sqrt{3}}$

$\csc \theta = \frac{r}{y} = \frac{2}{1} = 2$

$\sec \theta = \frac{r}{x} = - \frac{2}{\sqrt{3}}$

$\cot \theta = \frac{x}{y} = - \frac{\sqrt{3}}{1}$