# Given sinx=sqrt3/2, cosx=-1/2, how do you find the remaining trigonometric functions?

Jan 26, 2017

$\tan x = - \sqrt{3}$

$\sec x = - 2$

$\csc x = \frac{2}{3} \sqrt{3}$

$\cot x = - \frac{\sqrt{3}}{3}$

#### Explanation:

Since $\tan \theta = \sin \frac{\theta}{\cos} \theta$, you get

$\tan x = \frac{\frac{\sqrt{3}}{\cancel{2}}}{- \frac{1}{\cancel{2}}} = - \sqrt{3}$

Since $\sec \theta = \frac{1}{\cos} \theta$, you get

$\sec x = \frac{1}{- \frac{1}{2}} = - 2$

Since $\csc \theta = \frac{1}{\sin} \theta$, you get

$\csc x = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2}{3} \sqrt{3}$

Since $\cot \theta = \frac{1}{\tan} \theta$, you get

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