10. The sum of the maximum and minimum values of the function $y=\sin x+\sqrt{2+\cos ^{2} x}$ is $\qquad$ .
10. $2 \sqrt{2}$.

From $-1 \leqslant \sin x \leqslant 1, 0 \leqslant \cos ^{2} x \leqslant 1$, we know that when $x=\frac{3 \pi}{2}$, $\sin x+\sqrt{2+\cos ^{2} x}$ takes the minimum value $\sqrt{2}-1$.
Let $\sin x=\sqrt{3} \cos \theta$, where
$\theta \in\left[\arccos \frac{1}{\sqrt{3}}, \arccos \left(-\frac{1}{\sqrt{3}}\right)\right] \subset\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$.
Then $\sqrt{2+\cos ^{2} x}=\sqrt{3} \sin \theta$.
Thus, $y=\sqrt{6} \sin \left(\theta+\frac{\pi}{4}\right)$.
Since $\theta+\frac{\pi}{4} \in\left(\frac{\pi}{2}, \pi\right)$, we know that $y=\sqrt{6} \sin \left(\theta+\frac{\pi}{4}\right)$

is monotonically decreasing.
$$
\begin{array}{l}
\text { Then } y_{\text {max }}=\sqrt{6} \sin \left(\arccos \frac{1}{\sqrt{3}}+\frac{\pi}{4}\right) \\
=\sqrt{6}\left(\frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{2}}\right)=\sqrt{2}+1 .
\end{array}
$$

Then $y_{\text {max }}=\sqrt{6} \sin \left(\arccos \frac{1}{\sqrt{3}}+\frac{\pi}{4}\right)$

Therefore, the sum of the maximum and minimum values is $2 \sqrt{2}$.