11. Given real numbers $x, y$ satisfy: $1+\cos ^{2}(x+y-1)=\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1}$, find the minimum value of $x y$.

 untranslated part:
 已知实数 $x, y$ 满足: $1+\cos ^{2}(x+y-1)=\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1}$, 求 $x y$ 的最小值。

translated part:
 Given real numbers $x, y$ satisfy: $1+\cos ^{2}(x+y-1)=\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1}$, find the minimum value of $x y$.
Solve $1+\cos ^{2}(x+y-1)=\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1}=\frac{\left(x^{2}+y^{2}-2 x y\right)+2(x-y)+1+1}{x-y+1}$
$$
=\frac{(x-y+1)^{2}+1}{x-y+1}=x-y+1+\frac{1}{x-y+1} \text {. }
$$

Since $00$, thus $x-y+1+\frac{1}{x-y+1} \geq 2 \ldots \ldots \ldots \ldots$.
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array} { l } 
{ 1 + \operatorname { c o s } ^ { 2 } ( x + y - 1 ) = 2 } \\
{ x - y + 1 = 1 }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ \operatorname { c o s } ^ { 2 } ( x + y - 1 ) = 1 } \\
{ x = y }
\end{array} \Rightarrow \left\{\begin{array}{l}
x+y-1=k \pi, k \in \mathbf{Z} \\
x=y
\end{array}\right.\right.\right. \\
\Rightarrow x=y=\frac{1+k \pi}{2}, k \in \mathbf{Z} \Rightarrow x y=\left(\frac{1+k \pi}{2}\right)^{2} \geq \frac{1}{4}, k \in \mathbf{Z} \text {, hence }(x y)_{\min }=\frac{1}{4} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .20 \text { points } \\
\end{array}
$$