Official solution
\[{\log _{\cos x}}\cot x + 4{\log _{\sin x}}\tan x = 1\]
\[ \Rightarrow {\log _{\cos x}}\cot x - 4{\log _{\sin x}}\cot x = 1\]
\[ \Rightarrow 1 - {\log _{\cos x}}\sin x - 4 - 4{\log _{\sin x}}\cos x = 1\]
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 \[x \in \left( {0,{\pi \over 2}} \right)\]
\[\sin x = {{\sqrt 5 - 1} \over 2}\]
\[x = {\sin ^{ - 1}}\left( {{{ - 1 + \sqrt 5 } \over 2}} \right)\]
\[ \Rightarrow \alpha = - 1,\beta = 5\]
\[\alpha + \beta = 4\]