10. Eliminate $x, y$ from the following system of equations,
$$
\left\{\begin{array}{l}
\sin x+\sin y=a \\
\cos x+\cos y=b \\
\cot x \cot y=c
\end{array}\right.
$$
10. (1) $)^{2}+(2)^{2}$ we get $2+2(\sin x \sin y+\cos x \cos y)=a^{2}+b^{2}, \cos (x-y)=\frac{a^{2}+b^{2}}{2}-1$; $(2)^{2}$-(1) ${ }^{2}$ we get $\cos 2 x+\cos 2 y+2(\cos x \cos y-\sin x \sin y)=b^{2}-a^{2}, 2 \cos (x+y) \cos (x-y)+$ $2 \cos (x+y)=b^{2}-a^{2}, \cos (x+y)=\frac{b^{2}-a^{2}}{2(\cos (x-y)+1)}=\frac{b^{2}-a^{2}}{2\left(\frac{a^{2}+b^{2}}{2}-1+1\right)}=\frac{b^{2}-a^{2}}{b^{2}+a^{2}}$. From (3) we get $\frac{\cos x \cos y}{\sin x \sin y}=c \Rightarrow \frac{\cos (x+y)+\cos (x-y)}{\cos (x-y)-\cos (x+y)}=c$, so $\frac{\frac{b^{2}-a^{2}}{b^{2}+a^{2}}+\frac{a^{2}+b^{2}}{2}-1}{-\frac{b^{2}-a^{2}}{b^{2}+a^{2}}+\frac{b^{2}+a^{2}}{2}-1}=c$, which means $\frac{\left(a^{2}+b^{2}\right)^{2}-4 a^{2}}{\left(a^{2}+b^{2}\right)^{2}-4 b^{2}}=c$