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

Answer:

#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#