If $\sin A = n \sin B,$ then $\frac{n - 1}{n + 1} \tan \frac{A + B}{2} = $

Prove that $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$.

$\frac{\tan 80^{\circ}-\tan 10^{\circ}}{\tan 70^{\circ}}$ is equal to

$\cos^2 \left( \frac{\pi}{6} + \theta \right) - \sin^2 \left( \frac{\pi}{6} - \theta \right) = $

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

If $\alpha$ is in the $3^{\text{rd}}$ quadrant,$\beta$ is in the $2^{\text{nd}}$ quadrant such that $\tan \alpha = \frac{1}{7}$ and $\sin \beta = \frac{1}{\sqrt{10}}$,then find the value of $\sin(2\alpha + \beta)$.