Factorize the expression: $\sec ^2 x+5 \tan x+5$

If $x = \log \left[ \cot \left( \frac{\pi}{4} + \theta \right) \right]$, then the value of $\sinh x$ is

If $\cos \alpha = \frac{2 \cos \beta - 1}{2 - \cos \beta}$,then the value of $\tan \frac{\alpha}{2} \cot \frac{\beta}{2}$ is equal to,where $(0 < \alpha < \pi$ and $0 < \beta < \pi)$.

Evaluate the product: $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{2 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{4 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{6 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$

The value of $\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7}$ is

If $\cos A+\cos (A+B)+\cos (A+2 B)+\ldots$ up to $n$ terms $=$ $\cos \left(\frac{2 A+(n-1) B}{2}\right) \sin \frac{n B}{2} \operatorname{cosec} \frac{B}{2}$,then $\cos \frac{\pi}{19}+\cos \frac{3 \pi}{19}+\cos \frac{5 \pi}{19}+\ldots+\cos \frac{17 \pi}{19} = $