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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.
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