20. Find the sum of the maximum and minimum values of
$$
\frac{1}{1+(2 \cos x-4 \sin x)^{2}} \text {. }
$$
20. $\frac{22}{21}$
Note that
$$
2 \cos x-4 \sin x=\sqrt{20}\left(\frac{2}{\sqrt{20}} \cos x-\frac{4}{\sqrt{20}} \sin x\right) .
$$

Let $\varphi$ be a real number such that $\cos \varphi=\frac{2}{\sqrt{20}}$ and $\sin \varphi=\frac{4}{\sqrt{20}}$. We obtain
$$
2 \cos x-4 \sin x=\sqrt{20} \cos (x+\varphi) \text {. }
$$

Then
$$
0 \leq(2 \cos x-4 \sin x)^{2}=20 \cos ^{2}(x+\varphi) \leq 20 .
$$

Note here that we can particularly choose a value of $x$ so that $20 \cos ^{2}(x+$ $\varphi)=20$, and a value of $x$ so that $20 \cos ^{2}(x+\varphi)=0$. Furthermore, we get
$$
1 \leq 1+(2 \cos x-4 \sin x)^{2} \leq 21,
$$
and so
$$
\frac{1}{21} \leq \frac{1}{1+(2 \cos x-4 \sin x)^{2}} \leq 1 .
$$

Thus, the sum of the maximum and minimum values of the given expression is $1+\frac{1}{21}=\frac{22}{21}$.