You can observe that $\displaystyle \csc(90^o-\vartheta)=\sec(\vartheta)$. But if you didn't know that, how would you go about solving this problem?
Note 3 things:
$\displaystyle \sec(\varphi)=\frac{1}{\cos(\varphi)}$
$\displaystyle \csc(\varphi)=\frac{1}{\sin(\varphi)}$ and $\displaystyle \sin(\varphi-\vartheta)=\sin(\varphi)\cos(\vartheta)-\cos(\varphi)\sin(\vartheta)$
Thus,
$\displaystyle \csc(90^o-\vartheta)=\frac{1}{\sin(90^o-\vartheta)}=\frac{1}{\sin(90^o)\cos(\vartheta)-\cos(90^o)\sin(\vartheta)}$
Simplifying, we get:
$\displaystyle \csc(90^o-\vartheta)=\frac{1}{\cos(\vartheta)}=\sec(\varthet a)$
Since $\displaystyle \tan(\vartheta)=2$, we can conclude that $\displaystyle \cos(\vartheta)=\frac{1}{\sqrt{5}}$
Thus, $\displaystyle \csc(90^o-\vartheta)=\sec(\vartheta)=\color{red}\boxed{\sqrt {5}}$
Does this make sense?
--Chris