# How do you find the exact value of the five remaining trigonometric function given csctheta=-5 and the angle is in standard position in quadrant IV?

Nov 19, 2016

#### Explanation:

Given: $\csc \left(\theta\right) = - 5$

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

$\sin \left(\theta\right) = - \frac{1}{5}$

Use the identity $\cos \left(\theta\right) = \pm \sqrt{1 - {\sin}^{2} \left(\theta\right)}$ but in quadrant IV the cosine is positive so drop the negative:

$\cos \left(\theta\right) = \sqrt{1 - {\left(- \frac{1}{5}\right)}^{2}}$

$\cos \left(\theta\right) = \frac{2 \sqrt{6}}{5}$

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

$\sec \left(\theta\right) = \frac{5}{2 \sqrt{6}} = \frac{5 \sqrt{6}}{12}$

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

$\tan \left(\theta\right) = \frac{- \frac{1}{5}}{\frac{2 \sqrt{6}}{5}} = - \frac{\sqrt{6}}{12}$

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

$\cot \left(\theta\right) = \frac{\frac{2 \sqrt{6}}{5}}{- \frac{1}{5}} = - 2 \sqrt{6}$