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 $0 < \beta < \pi)$.

Let $\alpha, \beta$ and $\gamma$ be such that $0 < \alpha < \beta < \gamma < 2 \pi$. For any $x \in \mathbb{R}$,if $\cos (x+\alpha)+\cos (x+\beta)+\cos (x+\gamma)=0$,then $\tan (\gamma-\alpha) = $

Let $\alpha, \beta$ be two real numbers such that $\pi < (\alpha-\beta) < 3 \pi$. If $\sin \alpha+\sin \beta=\frac{-21}{65}$ and $\cos \alpha+\cos \beta=\frac{-27}{65}$,then $\cos \left(\frac{\beta-\alpha}{2}\right)=$

Prove that $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$.

If $\alpha, \beta$ are acute angles such that $\frac{\sin \alpha}{\sin \beta} = \frac{6}{5}$ and $\frac{\cos \alpha}{\cos \beta} = \frac{9}{5 \sqrt{5}}$,then $\sin \alpha = $

If $\sinh u = \tan \theta$,then $\cosh u$ is equal to