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

$A$ tree breaks at a certain height and the upper part touches the ground,making an angle of $60^{\circ}$ with the ground at a distance of $10 \text{ m}$ from its foot. The original height of the tree was (in $\text{m}$):

Let $A$ and $B$ denote the statements:
$A: \cos \alpha + \cos \beta + \cos \gamma = 0$
$B: \sin \alpha + \sin \beta + \sin \gamma = 0$
If $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$,then:

If $\cos \theta = \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$,then the value of $\cot \theta$ is equal to (where $0^{\circ} \leq \theta \leq 90^{\circ}$)

The value of $2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1$ is

If $\sqrt{2} \tan 2 \theta = \sqrt{6}$ and $0^{\circ} < \theta < 45^{\circ},$ then the value of $\sin \theta + \sqrt{3} \cos \theta - 2 \tan^{2} \theta$ is