# How do you find the remaining trigonometric functions of theta given cottheta=m/n where m and n are both positive?

Apr 2, 2017

$\tan \theta = \frac{n}{m}$

sintheta=(opposite)/(hypoten use)=n/(sqrt(n^2+m^2

costheta=(adjacent)/(hypoten use)=m/(sqrt(n^2+m^2)

$\csc \theta = \frac{1}{\sin} \theta = \frac{h y p o t e n u s e}{o p p o s i t e} = \frac{\sqrt{{n}^{2} + {m}^{2}}}{n}$

$\sec = \frac{1}{\cos} \theta = \frac{h y p o t e n u s e}{a \mathrm{dj} a c e n t} = \frac{\sqrt{{n}^{2} + {m}^{2}}}{m}$

#### Explanation:

First let's identify the ones that are left.

$\tan \theta , \sin \theta , \cos \theta , \sec \theta , \csc \theta$

But $\cot \theta = \frac{1}{\tan} \theta = \frac{a \mathrm{dj} a c e n t}{o p p o s i t e} = \frac{m}{n}$

$\tan \theta = \frac{o p p o s i t e}{a \mathrm{dj} a c e n t} = \frac{n}{m}$

Using the pythagorean theorem

$h y p o t e n u s {e}^{2} = o p p o s i t {e}^{2} + a \mathrm{dj} a c e n {t}^{2}$

$h y p o t e n u s e = \sqrt{o p p o s i t {e}^{2} + a \mathrm{dj} a c e n {t}^{2}}$

$h y p o t e n u s e = \sqrt{{n}^{2} + {m}^{2}}$

sintheta=(opposite)/(hypoten use)=n/(sqrt(n^2+m^2

costheta=(adjacent)/(hypoten use)=m/(sqrt(n^2+m^2)

$\csc \theta = \frac{1}{\sin} \theta = \frac{h y p o t e n u s e}{o p p o s i t e} = \frac{\sqrt{{n}^{2} + {m}^{2}}}{n}$

$\sec = \frac{1}{\cos} \theta = \frac{h y p o t e n u s e}{a \mathrm{dj} a c e n t} = \frac{\sqrt{{n}^{2} + {m}^{2}}}{m}$