1. We need to evaluate

$$\sin 70^\circ\left(\cot 10^\circ\cot 70^\circ-1\right).$$

2. Use the identity

$$\cot 70^\circ = \tan 20^\circ$$

because

$$\cot \theta = \tan(90^\circ-\theta).$$

So the expression becomes

$$\sin 70^\circ\left(\cot 10^\circ\tan 20^\circ-1\right).$$

3. Also,

$$\sin 70^\circ = \cos 20^\circ.$$

Hence

$$\sin 70^\circ\left(\cot 10^\circ\cot 70^\circ-1\right) = \cos 20^\circ\left(\cot 10^\circ\tan 20^\circ-1\right).$$

4. Now write in terms of sine and cosine:

$$\cot 10^\circ\tan 20^\circ = \frac{\cos 10^\circ}{\sin 10^\circ}\cdot \frac{\sin 20^\circ}{\cos 20^\circ}.$$

Using

$$\sin 20^\circ = 2\sin 10^\circ\cos 10^\circ,$$

we get

$$\cot 10^\circ\tan 20^\circ = \frac{\cos 10^\circ}{\sin 10^\circ}\cdot \frac{2\sin 10^\circ\cos 10^\circ}{\cos 20^\circ} = \frac{2\cos^2 10^\circ}{\cos 20^\circ}.$$

5. Therefore,

$$\cos 20^\circ\left(\cot 10^\circ\tan 20^\circ-1\right) = \cos 20^\circ\left(\frac{2\cos^2 10^\circ}{\cos 20^\circ}-1\right) = 2\cos^2 10^\circ-\cos 20^\circ.$$

6. Use the identity

$$\cos 20^\circ = 2\cos^2 10^\circ-1.$$

So

$$2\cos^2 10^\circ-\cos 20^\circ = 2\cos^2 10^\circ-(2\cos^2 10^\circ-1)=1.$$

7. Hence the value is

$$\boxed{1}.$$

8. Checking options:

• A: $0$ ❌
• B: $\frac{2}{3}$ ❌
• C: $1$ ✅
• D: $\frac{3}{2}$ ❌

So the correct option is C.