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

1 Answer
Apr 2, 2017

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