# How do you find the exact value of 6 trig function given a point (-sqrt5, 2)?

Nov 18, 2016

#### Explanation:

Use the identity: $\tan \left(\theta\right) = \frac{y}{x}$:

$\tan \left(\theta\right) = \frac{2}{-} \sqrt{5}$

Rationalize the denominator:

$\tan \left(\theta\right) = \frac{- 2 \sqrt{5}}{5}$

Use the identity: $\cot \left(\theta\right) = \frac{1}{\tan} \left(\theta\right)$

$\cot \left(\theta\right) = - \frac{\sqrt{5}}{2}$

Use the identity ${\tan}^{2} \left(\theta\right) + 1 = {\sec}^{2} \left(\theta\right)$

$\frac{4}{5} + 1 = {\sec}^{2} \left(\theta\right)$

${\sec}^{2} \left(\theta\right) = \frac{9}{5}$

$\sec \left(\theta\right) = \pm \frac{3}{\sqrt{5}}$

Rationalize the denominator:

$\sec \left(\theta\right) = \pm \frac{3 \sqrt{5}}{5}$

Please observe that the point $\left(- \sqrt{5} , 2\right)$ is in the second quadrant and the secant is negative in the second quadrant, therefore, we drop the + sign:

$\sec \left(\theta\right) = \frac{- 3 \sqrt{5}}{5}$

Use the identity $\cos \left(\theta\right) = \frac{1}{\sec} \left(\theta\right)$:

$\cos \left(\theta\right) = - \frac{\sqrt{5}}{3}$

Use the identity $\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right)$:

$\sin \left(\theta\right) = \tan \left(\theta\right) \cos \left(\theta\right)$

$\sin \left(\theta\right) = \left(\frac{2}{-} \sqrt{5}\right) \left(- \frac{\sqrt{5}}{3}\right)$

$\sin \left(\theta\right) = \frac{2}{3}$

Use the identity: $\csc \left(\theta\right) = \frac{1}{\sin} \left(\theta\right)$:

$\csc \left(\theta\right) = \frac{3}{2}$