The points #(2/3,5/2)# and #(4,15)# fall on the same line and therefore the terminal side of the same angle, #theta#, and the second point is easier to work with. (I basically multiplied through by 6 to scale the point up.)
Calculate #r=sqrt(4^2+15^2)=sqrt(16+225)=sqrt(241)#.
Knowing #x#, #y#, and #r# we can find:
#cos(theta)=x/r#, #sin(theta)=y/r#, and #tan(theta)=y/x#.
For this problem we have:
#cos(theta)=4/sqrt(241)#, #sin(theta)=15/sqrt(241)#, and #tan(theta)=15/4#.
The remaining functions are found by taking reciprocals:
#sec(theta)=sqrt(241)/4#, #csc(theta)=sqrt(241)/15#, and #cot(theta)=4/15#.
