# How do you use fundamental identities to find the values of the trigonometric values given sin X = - 5/13 and tan X > 0?

Feb 8, 2017

$\sin x = - \frac{5}{13}$, $\cos x = - \frac{12}{13}$, $\tan x = \frac{5}{12}$

$\cot x = \frac{12}{5}$, $\sec x = - \frac{13}{12}$, $\csc x = - \frac{13}{5}$

#### Explanation:

As $\sin x = - \frac{5}{13}$ and $\tan x > 0$ i.e. positive,

$\cos x$ which is $\sin \frac{x}{\tan} x$ is negative.

Hence $\cos x = - \sqrt{1 - {\sin}^{2} x} = - \sqrt{1 - {\left(- \frac{5}{13}\right)}^{2}}$

= $- \sqrt{1 - \frac{25}{169}} = - \sqrt{\frac{144}{169}} = - \frac{12}{13}$

and $\tan x = \sin \frac{x}{\cos} x = \frac{- \frac{5}{13}}{- \frac{12}{13}} = - \frac{5}{13} \times - \frac{13}{12} = \frac{5}{12}$

$\cot x = \frac{1}{\tan} x = \frac{12}{5}$

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

$\csc x = \frac{1}{\sin} x = - \frac{13}{5}$