We can use 3 facts:
[tex] \tan{\theta}=\frac{y}{x} [/tex]
[tex] \sec{\theta}=\frac{r}{x} [/tex]
[tex] x^2+y^2=r^2 [/tex]
So, to find our x and our y we can do:
[tex]
\tan^2{\theta}=\frac{3}{8} \implies \\
\tan{\theta}=\pm{\sqrt{\frac{3}{8}}=\pm\frac{\sqrt{3}}{\sqrt{8}}
[/tex]So, now that we know our x and our y we can find our r.
[tex] (\sqrt{3})^2+(\sqrt{8})^2=11 \implies r=\sqrt{11} [/tex]
You'll notice, though, that there were two solutions for [tex] \tan{\theta} [/tex], which means we have a positive and a negative for our x. So:
[tex] \sec{\theta}=\frac{\sqrt{11}}{\sqrt{8}} \\
\sec{\theta}=-\frac{\sqrt{11}}{\sqrt{8}}}\\
\text{Rationalizing both of them:}\\
\sec{\theta}=\frac{\sqrt{88}}{8} \\
\sec{\theta}=-\frac{\sqrt{88}}{8} [/tex]
