# How do you find the remaining trigonometric functions of theta given csctheta=13/5 and costheta<0?

Apr 14, 2017

$\sin \theta = \frac{5}{13}$, $\cos \theta = - \frac{12}{13}$, $\tan \theta = - \frac{5}{12}$, $\cot \theta = - \frac{12}{5}$,
$\sec \theta = - \frac{13}{12}$;and $\csc \theta = \frac{13}{5}$

#### Explanation:

As $\csc \theta = \frac{13}{5}$, it is positive and as $\cos \theta$ is negative, $\theta$ lies in Quadrant II.

Now $\sin \theta = \frac{1}{\csc} \theta = \frac{1}{\frac{13}{5}} = \frac{5}{13}$

$\cos \theta = - \sqrt{1 - {\left(\frac{5}{13}\right)}^{2}} = - \sqrt{1 - \frac{25}{169}} = - \sqrt{\frac{144}{169}} = - \frac{12}{13}$

tantheta=sintheta/costheta=(5/13)/(-12/13)=-5/13×13/12=-5/12

$\cot \theta = \frac{1}{\tan} \theta = \frac{1}{- \frac{5}{12}} = - \frac{12}{5}$

$\sec \theta = \frac{1}{\cos} \theta = \frac{1}{- \frac{12}{13}} = - \frac{13}{12}$