$$
\begin{array}{l}
\text { 20. Let } 135^{\circ}<\alpha<180^{\circ} \text {, and } \\
P\left(\cos \alpha, \cos ^{2} \alpha\right), Q\left(\cot \alpha, \cot ^{2} \alpha\right), \\
R\left(\sin \alpha, \sin ^{2} \alpha\right), S\left(\tan \alpha, \tan ^{2} \alpha\right)
\end{array}
$$

be the four vertices of a trapezoid. Then $\sin 2 \alpha=(\quad)$.
(A) $2-2 \sqrt{2}$
(B) $3 \sqrt{3}-6$
(C) $3 \sqrt{2}-5$
(D) $-\frac{3}{4}$
(E) $1-\sqrt{3}$
20. A.

Let $f, g, h, j$ represent the sine, cosine, tangent, and cotangent functions (in no particular order). Then the four vertices of the trapezoid are $\left(f, f^{2}\right),\left(g, g^{2}\right),\left(h, h^{2}\right),\left(j, j^{2}\right)$.
Since quadrilateral $P Q R S$ is a trapezoid, the two bases are parallel, meaning their slopes are equal. Without loss of generality, we have
$$
\frac{g^{2}-f^{2}}{g-f}=\frac{j^{2}-h^{2}}{j-h} \Rightarrow g+f=j+h .
$$

Thus, there are three possible scenarios:
(1) $\sin \alpha+\cos \alpha=\tan \alpha+\cot \alpha$;
(2) $\sin \alpha+\tan \alpha=\cos \alpha+\cot \alpha$;
(3) $\sin \alpha+\cot \alpha=\cos \alpha+\tan \alpha$.

Given $135^{\circ}<\alpha<180^{\circ}$, we know
$0<\sin \alpha<\frac{\sqrt{2}}{2},-1<\cos \alpha<-\frac{\sqrt{2}}{2}$,
$-1<\tan \alpha<0, \cot \alpha<-1$,
which implies $\cot \alpha<\cos \alpha<\sin \alpha, \cot \alpha<\tan \alpha<\sin \alpha$.
Thus, it is easy to verify that scenarios (1) and (2) do not satisfy the conditions.
Next, we prove that scenario (3) holds.
From $\sin \alpha+\frac{\cos \alpha}{\sin \alpha}=\cos \alpha+\frac{\sin \alpha}{\cos \alpha}$
$\Rightarrow(\sin \alpha-\cos \alpha)\left(1-\frac{\sin \alpha+\cos \alpha}{\sin \alpha \cdot \cos \alpha}\right)=0$.
Since $\sin \alpha \neq \cos \alpha$, we have
$\sin \alpha+\cos \alpha=\sin \alpha \cdot \cos \alpha$
$\Rightarrow(\sin \alpha \cdot \cos \alpha)^{2}-2 \sin \alpha \cdot \cos \alpha-1=0$.
Given that $\sin \alpha \cdot \cos \alpha<0$, we get $\sin \alpha \cdot \cos \alpha=1-\sqrt{2}$.
Therefore, $\sin 2 \alpha=2 \sin \alpha \cdot \cos \alpha=2-2 \sqrt{2}$.