Due to periodic properties of sine and cosine we have $\cot730^{\circ} = \frac{1}{\tan730^{\circ}} = \cot10^{\circ} = \frac{1}{\tan10^{\circ}}$ and $\cot800^{\circ} = \frac{1}{\tan800^{\circ}} = \cot80^{\circ} = \frac{1}{\tan80^{\circ}}$. Therefore:
$$
\frac{1}{\cot730^{\circ}\cot800^{\circ}+\tan730^{\circ}\tan800^{\circ}}
=
\frac{1}{\cot10^{\circ}\cot80^{\circ}+\tan10^{\circ}\tan80^{\circ}}
$$
Now note that
$$\cos10^{\circ} = \cos(-10^{\circ}) = \sin(-10^{\circ} + 90^{\circ}) = \sin80^{\circ},$$
$$\cos80^{\circ} = -\sin(80^{\circ}-90^{\circ})= -\sin(-10^{\circ}) = \sin10^{\circ},$$
so
$$
\tan10^{\circ} = \frac{\sin10^{\circ}}{\cos10^{\circ}}= \frac{\cos80^{\circ}}{\sin80^{\circ}} = \cot80^{\circ},
$$
$$
\tan80^{\circ} = \frac{\sin80^{\circ}}{\cos80^{\circ}}= \frac{\cos10^{\circ}}{\sin10^{\circ}} = \cot10^{\circ},
$$
so you have
$$
\frac{1}{\cot10^{\circ}\cot80^{\circ}+\tan10^{\circ}\tan80^{\circ}}
=
\frac{1}{\cot10^{\circ}\tan10^{\circ}+\tan10^{\circ}\cot10^{\circ}}
=
\frac{1}{2}
$$