10. Given the function
$$
f(x)=\cos (a \sin x)-\sin (b \cos x)
$$

has no zeros. Then the range of $a^{2}+b^{2}$ is ( ).
(A) $\left[0, \frac{\pi}{4}\right)$
(B) $\left[0, \frac{\pi^{2}}{2}\right)$
(C) $\left[0, \frac{\pi^{2}}{4}\right)$
(D) $\left[0, \frac{\pi}{2}\right)$
10. C.

From the given, we have
$$
\left\{\begin{array}{l}
2 k \pi + b \cos x = \frac{\pi}{2} - a \sin x, \\
(2 k + 1) \pi - b \cos x = \frac{\pi}{2} - a \sin x
\end{array} \quad (k \in \mathbf{Z})\right.
$$

has no solution.
Then $\sqrt{a^{2}+b^{2}} \sin (x+\varphi) = \frac{\pi}{2} + 2 k \pi$ or
$$
\sqrt{a^{2}+b^{2}} \sin (x-\varphi) = \frac{\pi}{2} - (2 k + 1) \pi,
$$

where $\sin \varphi = \frac{b}{\sqrt{a^{2}+b^{2}}}, \cos \varphi = \frac{a}{\sqrt{a^{2}+b^{2}}}$, has no solution.
Therefore, $\sqrt{a^{2}+b^{2}} < \min \left\{\left|\frac{\pi}{2} + 2 k \pi\right|, \left|\frac{\pi}{2} - (2 k + 1) \pi\right|\right\}$ $\Rightarrow$ the range of $a^{2}+b^{2}$ is $\left[0, \frac{\pi^{2}}{4}\right)$.