As sin(theta)>0, then theta terminates in either QI or QII.
If theta terminates in QI, then cos(theta)>0. By the Pythagorean identity sin^2(theta)+cos^2(theta)=1, we have (3sqrt(13)/13)^2+cos^2(theta)=1. Hence cos^2(theta) = 4/13. Therefore cos(theta)=2sqrt(13)/13 as cos(theta)>0 in this case. So sec(theta)=1/cos(theta)=sqrt(13)/2.
If theta terminates in QII, then cos(theta)<0. By the Pythagorean identity sin^2(theta)+cos^2(theta)=1, we have (3sqrt(13)/13)^2+cos^2(theta)=1. Hence cos^2(theta) = 4/13. Therefore cos(theta)=-2sqrt(13)/13 as cos(theta)<0 in this case. So sec(theta)=1/cos(theta)=-sqrt(13)/2.
Now we conclude that sec(theta)=sqrt(13)/2 if theta terminates in QI and sec(theta)=-sqrt(13)/2 if theta terminates in QII.
