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

${\log }_{\mathrm{cosx}}\frac{\mathrm{cosx}}{\mathrm{sinx}}+4.{\log }_{\mathrm{sinx}}\frac{\mathrm{sinx}}{\mathrm{cosx}}=1$

$1-{\log }_{\cos x} \sin x+4-4{\log }_{\sin x} \cos x=1$

${\log }_{\cos x} \sin x+4{\log }_{\sin x} \cos x=4$

$\Rightarrow \mathrm{t}+\frac{4}{\mathrm{t}}=4\Rightarrow \mathrm{t}=2 \mathrm{where} \mathrm{t}={\log }_{\mathrm{cosx}}^{\mathrm{sinx}}$

${\log }_{\mathrm{cosx}}^{\mathrm{sinx}}=2\Rightarrow \mathrm{sinx}={\cos }^2\mathrm{x}$

${\sin }^{2}x+\sin x-1=0$

$\Rightarrow \mathrm{sinx}=\frac{-1+\sqrt{5}}{2}=\frac{\mathrm{\alpha }+\sqrt{\mathrm{\beta }}}{2}$

$\mathrm{\alpha }\mathrm{+}\mathrm{\beta }=-1+5=4$