If $m \cos (\alpha+\beta)-n \cos (\alpha-\beta)=m \cos (\alpha-\beta)+n \cos (\alpha+\beta)$,then $\tan \alpha \tan \beta=$

Prove that: $2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} + \cos \frac{3 \pi}{13} + \cos \frac{5 \pi}{13} = 0$

Let $\cos(\alpha+\beta)=-\frac{1}{10}$ and $\sin(\alpha-\beta)=\frac{3}{8}$ where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha=\frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}$, where $r, s \in N$, then $r+s$ is equal to . . . . . .

$\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B} = $

$\tan 20^\circ + \tan 40^\circ + \sqrt{3} \tan 20^\circ \tan 40^\circ = $

The value of $\cos 15^{\circ} - \sin 15^{\circ}$ is ......