$\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} = $

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

The value of $\left(\sin 70^{\circ}\right)\left(\cot 10^{\circ} \cot 70^{\circ}-1\right)$ is

If $\sec \theta + \tan \theta = 2/3$,then in which quadrant does $\theta$ lie?

$\tan 2 \alpha \cdot \tan \left(30^{\circ}-\alpha\right)+\tan 2 \alpha \cdot \tan \left(60^{\circ}-\alpha\right)+\tan \left(60^{\circ}-\alpha\right) \cdot \tan \left(30^{\circ}-\alpha\right)$ is equal to

If $\cos \theta - 4 \sin \theta = 1$,then $\sin \theta + 4 \cos \theta$ is equal to