Trigonometry

We can solve the problem using Weierstrass or tangent half-angle substitution which gives that $\cos x = \dfrac {1-t^2}{1+t^2}$ , where $t = \tan \frac x2$ . Let $a = \tan \frac \alpha 2$ and $b = \tan \frac \beta 2$ ; then $\tan \frac \alpha 2 \cot \frac \beta 2 = \dfrac ab$ .

Now we have:

$\begin{aligned} \cos \alpha & = \frac {2 \cos \beta -1}{2 - \cos \beta} \\ \frac {1-a^2}{1+a^2} & = \frac {2\left(\frac {1-b^2}{1+b^2}\right)-1}{2-\frac {1-b^2}{1+b^2}} \\ & = \frac {2-2b^2-1-b^2}{2+2b^2-1+b^2} \\ & = \frac {1-3b^2}{1+3b^2} \\ (1-a^2)(1+3b^2) & = (1+a^2)(1-3b^2) \\ 1-a^2+3b^2-3a^2b^2 & = 1+a^2-3b^2-3a^2b^2 \\ 6b^2 & = 2a^2 \\ \frac {a^2}{b^2} & = 3 \\ \implies \frac ab = \tan \frac \alpha 2 \cot \frac \beta 2 & = \sqrt 3 \approx \boxed{1.732} \end{aligned}$

There must be a much shorter solution. Anyway i solved it in the following way

To find $\large{\tan\left(\frac{\alpha}{2}\right) \times \cot\left(\frac{\beta}{2}\right)}$ or we can find

$\large{\frac{1- \cos \alpha}{sin \alpha}\times \frac{\sin \beta}{1-\cos \beta}}$

Firstly

$\large{\Rightarrow 1-cos \alpha=1-\left(\frac{ 2 \cos \beta-1}{2-\cos \beta}\right)}$

$\large{\Rightarrow 1-\cos \alpha=\frac{3(1-\cos \beta)}{2-\cos \beta}...........(1)}$

Now to find $\sin \alpha$

$\large{\sqrt{1-\sin^{2} \alpha}= \frac{ 2 \cos \beta-1}{2-\cos \beta}}$

On squaring and simplifying

$\large{\sin \alpha= \frac{\sqrt{3}(\sin \beta)}{2-\cos \beta}....................(2)}$

Note: I took the positive value because that matches with the answer.

Apply $(1)/(2)$ and simplify to get

$\large{\frac{1- \cos \alpha}{sin \alpha}\times \frac{\sin \beta}{1-\cos \beta}=\sqrt{3}}$