If $$\sin \theta=-\frac{12}{13}, \cos \phi=-\frac{4}{5}$$ and $$\theta, \phi$$ lie in the third quadrant, then $$\tan (\theta-\phi)=$$

$$\frac{1-2\left[\cos 60^{\circ}-\cos 80^{\circ}\right]}{2 \sin 10^{\circ}}=\ldots \ldots$$

The value of $\sin 18^{\circ}$ is $\qquad$

If $\theta=\frac{17 \pi}{3}$ then, $\tan \theta-\cot \theta=\ldots$