${\log _{\cos x}}\cot x + 4{\log _{\sin x}}\tan x = 1$ $\Rightarrow {\log _{\cos x}}\cot x - 4{\log _{\sin x}}\cot x = 1$ $\Rightarrow 1 - {\log _{\cos x}}\sin x - 4 - 4{\log _{\sin x}}\cos x = 1$ Let ${\log _{\cos x}}\sin x = t$ $t + {4 \over t} = 4$ $\Rightarrow t = 2$ $\sin x = {\cos ^2}x$ $\Rightarrow \sin x = 1 - {\sin ^2}x$ $\Rightarrow {\sin ^2}x + \sin {x^{ - 1}} = 0$ $\Rightarrow \sin x = {{ - 1\, \pm \,\sqrt 5 } \over 2}$ as $x \in \left( {0,{\pi \over 2}} \right)$ $\sin x = {{\sqrt 5 - 1} \over 2}$ $x = {\sin ^{ - 1}}\left( {{{ - 1 + \sqrt 5 } \over 2}} \right)$ $\Rightarrow \alpha = - 1,\beta = 5$ $\alpha + \beta = 4$