# How do you find the exact value of the five remaining trigonometric function given costheta=3/5 and the angle is in standard position in quadrant IV?

Nov 10, 2016

#### Explanation:

Given: $\cos \left(\theta\right) = \frac{3}{5}$ and $\theta$ is in quadrant IV

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

$\sin \left(\theta\right) = \pm \sqrt{1 - {\left(\frac{3}{5}\right)}^{2}}$

$\sin \left(\theta\right) = \pm \sqrt{1 - \frac{9}{25}}$

$\sin \left(\theta\right) = \pm \sqrt{\frac{16}{25}}$

$\sin \left(\theta\right) = \pm \frac{4}{5}$

The sine function is negative in quadrant IV, therefore, we change the $\pm$ to only $-$:

$\sin \left(\theta\right) = - \frac{4}{5}$

$\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right)$

$\tan \left(\theta\right) = \frac{- \frac{4}{5}}{\frac{3}{5}}$

$\tan \left(\theta\right) = - \frac{4}{3}$

$\cot \left(\theta\right) = \frac{1}{\tan} \left(\theta\right)$

$\cot \left(\theta\right) = - \frac{3}{4}$

$\sec \left(\theta\right) = \frac{1}{\cos} \left(\theta\right)$

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

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

$\csc \left(\theta\right) = - \frac{5}{4}$