If  the solution of the equation  ${\log }_{\cos x}\cot x+4{\log }_{\sin x}\tan x=1,x\in (0,\frac{\pi }{2})$ , is  ${\sin }^{-1}\left(\frac{\alpha +\sqrt{\beta }}{2}\right)$, where $\alpha ,\beta$  are integers, then $\alpha +\beta$ is equal to:

$$\begin{gathered} {\log }_{\cos x}\cot x+4{\log }_{\sin x}\tan x=1 \\ \Rightarrow \frac{\mathrm{lncos}x-\mathrm{lnsin}x}{\mathrm{lncos}x}+4\frac{\mathrm{lnsin}x-\mathrm{lncos}x}{\mathrm{lnsin}x} \\ \Rightarrow {\left(\mathrm{lnsin}x\right)}^2-4\left(\mathrm{lnsin}x\right)\left(\mathrm{lncos}x\right)+4{\left(\mathrm{lncos}x\right)}^2=1 \\ \Rightarrow \mathrm{lnsin}x=2\mathrm{lncos}x \\ \Rightarrow {\sin }^{2}x+\sin x-1=0\Rightarrow \sin x=\frac{-1+\sqrt{5}}{2} \\ \therefore \alpha +\beta =4 \end{gathered}$$