If $\tan A = \frac{1 - \cos B}{\sin B},$ find $\tan 2A$ in terms of $\tan B$.

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} = $

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

If $\theta$ is an acute angle and $\sin \frac{\theta}{2} = \sqrt{\frac{x - 1}{2x}}$,then $\tan \theta$ is equal to

The value of $\tan \left(\frac{7 \pi}{8}\right)$ is

If $a \sin \theta = b \cos \theta$,where $a, b \neq 0$,then $a \cos 2 \theta + b \sin 2 \theta = $