1. In $\triangle A B C$, $a, b, c$ are the sides opposite to $\angle A, \angle B, \angle C$ respectively, satisfying $a^{2}+b^{2}=4-\cos ^{2} C, a b=2$. Then $S_{\triangle A B C}=$ $\qquad$
$-1.1$
From the problem, we have $(a-b)^{2}+\cos ^{2} C=0$.
Solving, we get $a=b=\sqrt{2}, \cos C=0$.
Therefore, $S_{\triangle A B C}=1$.