$\tan \alpha + 2 \tan 2 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha = $

If $A$ is not an integral multiple of $\frac{\pi}{2}$,then $\operatorname{cosec} 2A + \cot 2A$ is equal to

$\cos A \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$ equals

If $\tan \frac{\theta}{2} = t$,then $\frac{1 - t^2}{1 + t^2}$ is equal to

$\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = $ (when $x$ lies in $II^{nd}$ quadrant)

If $\frac{\sec 8\theta - 1}{\sec 4\theta - 1} = \frac{a + b \tan^2 2\theta}{1 + c \tan^2 2\theta + d \tan^4 2\theta}$ (where $\theta \neq \frac{n\pi}{16}, n \in I$),then the value of $(a - b + c - d)$ is -