If $\tan B = \frac{n \sin A \cos A}{1 - n \cos^2 A}$,then $\tan(A + B)$ equals

$\cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} - \cos \frac{4\pi}{7} + \cos \frac{5\pi}{7} - \cos \frac{6\pi}{7} = $

If $\tan \alpha = \frac{x^2 - x}{x^2 - x + 1}$ and $\tan \beta = \frac{1}{2x^2 - 2x + 1}$ $(x \ne 0, 1)$,where $0 < \alpha, \beta < \frac{\pi}{2}$,then $\tan(\alpha + \beta)$ has the value equal to:

If $\tan \beta = \frac{\tan \alpha + \tan \gamma}{1 + \tan \alpha \tan \gamma}$,then $\frac{\sin 2 \alpha + \sin 2 \gamma}{1 + \sin 2 \alpha \sin 2 \gamma} = $

Let $\theta, \phi \in [0, 2\pi]$ be such that $2 \cos \theta(1-\sin \phi) = \sin^2 \theta \left(\tan \frac{\theta}{2} + \cot \frac{\theta}{2}\right) \cos \phi - 1$,$\tan (2\pi - \theta) > 0$ and $-1 < \sin \theta < -\frac{\sqrt{3}}{2}$. Then $\phi$ cannot satisfy

If $\frac{3\pi}{4} < \alpha < \pi,$ then $\sqrt{\csc^2 \alpha + 2\cot \alpha}$ is equal to