The value of $$\left(\sin 70^{\circ}\right)\left(\cot 10^{\circ}\cot 70^{\circ}-1\right)$$ is

We need to find the value of $$(\sin 70°)(\cot 10° \cot 70° - 1)$$.

$$\sin 70° \left(\frac{\cos 10° \cos 70°}{\sin 10° \sin 70°} - 1\right) = \sin 70° \cdot \frac{\cos 10° \cos 70° - \sin 10° \sin 70°}{\sin 10° \sin 70°}$$

$$\cos 10° \cos 70° - \sin 10° \sin 70° = \cos(10° + 70°) = \cos 80°$$.

$$= \sin 70° \cdot \frac{\cos 80°}{\sin 10° \sin 70°} = \frac{\cos 80°}{\sin 10°}$$

Since $$\cos 80° = \sin 10°$$:

$$= \frac{\sin 10°}{\sin 10°} = 1$$

The answer is $$\boxed{1}$$, which corresponds to Option 2.