10. Find the range of the function $y=\frac{x^{2}+2 x \cdot \sin \theta+2}{x^{2}+2 x \cdot \cos \theta+2}$ ( $\theta$ is a parameter).
10. When $x=0$, $y=1$, when $x \neq 0$, make the substitution $t=\frac{x^{2}+2}{2 x}$, then $|t|=\left|\frac{x}{2}+\frac{1}{x}\right|=\frac{|x|}{2}+\frac{1}{|x|} \geqslant \sqrt{2}, y=$ $\frac{t+\sin \theta}{t+\cos \theta}$, that is, $y t-1=\sin \theta-y \cdot \cos \theta$, construct a circle with diameter $AB=1$, construct chord $AC=|\sin \theta|$, construct chord $AD=\frac{|y|}{\sqrt{1+y^{2}}}$, then $BD$ $=\frac{1}{\sqrt{1+y^{2}}}, BC=|\cos \theta|$. By Ptolemy's theorem and $CD \leqslant AB=1$, we have $|\sin \theta|+|y \cos \theta| \leqslant \sqrt{1+y^{2}}$, also $|y t-t|=|\sin \theta-y \cos \theta| \leqslant|\sin \theta|+|y \cos \theta| \leqslant \sqrt{1+y^{2}}$, that is, $|t| \cdot|y-1| \leqslant \sqrt{1+y^{2}}, \sqrt{2}|y-1| \leqslant$ $\sqrt{1+y^{2}}$, hence $2-\sqrt{3} \leqslant y \leqslant 2+\sqrt{3}$.