10. If $a^{2}+b^{2}=1, c^{2}+d^{2}=4$, then the range of values for $a d+b c$ is
10. $[-2,2]$.

Let $a=\cos \alpha$, then $b=\sin \alpha, c=2 \cos \beta$, and $d=2 \sin \beta, \alpha \in[0,2 \pi], \beta \in[0,2 \pi]$. Therefore, $a d + b c = 2 \cos \alpha \sin \beta + 2 \sin \alpha \cos \beta = 2 \sin (\alpha + \beta)$, hence $-2 \leqslant a d + b c \leqslant 2$.