For the value of $\frac{2 \tan(x)}{1-\tan^2(x)}$ to be positive,find the values of $x$ such that $x \in \left(0, \frac{\pi}{2}\right)$.

Let $0 < x < \frac{\pi}{4}.$ Then $\sec 2x - \tan 2x = $

If $\tan \alpha = \frac{-12}{5}$,$\cot \beta = \frac{7}{24}$,$\alpha$ does not belong to the second quadrant,and $\beta$ does not belong to the first quadrant,then $\sqrt{13} \sin \frac{\alpha}{2} + \cos \frac{\beta}{2} + \tan \frac{\alpha}{2} \cot \frac{\beta}{2} = $

If $\cos A = \frac{\sqrt{3}}{2},$ then $\tan 3A = $

$\frac{1-\tan^2 15^{\circ}}{1+\tan^2 15^{\circ}} = ?$

$\tan \left(\frac{3 \pi}{16}\right)+\cot \left(\frac{3 \pi}{16}\right)=$