Question

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

Answer 1

`tantheta=n/m`

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

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

`csctheta=1/sintheta=(hypoten use)/(opposite)=(sqrt(n^2+m^2))/n`

`sec=1/costheta=(hypoten use)/(adjacent)=sqrt(n^2+m^2)/m`

Explanation:

First let's identify the ones that are left.

`tantheta, sintheta,costheta,sectheta,csctheta`

But `cottheta=1/tantheta=(adjacent)/(opposite)=m/n`

`tantheta=(opposite)/(adjacent)=n/m`

Using the pythagorean theorem

`hypoten use^2=opposite^2+adjacent^2`

`hypoten use=sqrt(opposite^2+adjacent^2)`

`hypoten use=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)`

`csctheta=1/sintheta=(hypoten use)/(opposite)=(sqrt(n^2+m^2))/n`

`sec=1/costheta=(hypoten use)/(adjacent)=sqrt(n^2+m^2)/m`