# If sin theta = -5/13 and cos theta = -12/13, how do you find cot theta?

Apr 1, 2018

$\cot \theta = \frac{12}{5}$

#### Explanation:

We know that:

$\tan \theta = \sin \frac{\theta}{\cos} \theta$

and

$\cot \theta = \frac{1}{\tan} \theta$

So we can plug in our values:

$\textcolor{w h i t e}{=} \cot \left(\theta\right)$

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

$= \frac{1}{\sin \frac{\theta}{\cos} \theta}$

$= \frac{1}{\sin \frac{\theta}{\cos} \theta} \textcolor{b l u e}{\cdot \cos \frac{\theta}{\cos} \theta}$

$= \frac{1}{\sin \frac{\theta}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\cos}}}} \theta} \textcolor{b l u e}{\cdot \cos \frac{\theta}{\textcolor{red}{\cancel{\textcolor{b l u e}{\cos}}}} \theta}$

$= \frac{1}{\sin} \theta \textcolor{b l u e}{\cdot \cos \theta}$

$= \cos \frac{\theta}{\sin} \theta$

Now plug in our values:

$= \frac{- \frac{12}{13}}{- \frac{5}{13}}$

$= \frac{- \frac{12}{13}}{- \frac{5}{13}} \textcolor{b l u e}{\cdot \frac{13}{13}}$

$= \frac{- \frac{12}{\textcolor{red}{\cancel{\textcolor{b l a c k}{13}}}}}{- \frac{5}{\textcolor{red}{\cancel{\textcolor{b l a c k}{13}}}}} \textcolor{b l u e}{\cdot \frac{\textcolor{red}{\cancel{\textcolor{b l u e}{13}}}}{\textcolor{red}{\cancel{\textcolor{b l u e}{13}}}}}$

$= \frac{- 12}{- 5} \textcolor{b l u e}{\cdot \frac{1}{1}}$

$= \frac{- 12}{- 5}$

$= \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{-}}} 12}{\textcolor{red}{\cancel{\textcolor{b l a c k}{-}}} 5}$

$= \frac{12}{5}$

That's the solution. Hope this helped!