Official solution
As we know,
\[{\cos ^{ - 1}}A - {\cos ^{ - 1}}B\]
\[ = {\cos ^{ - 1}}\left( {AB + \sqrt {1 - {A^2}} .\sqrt {1 - {B^2}} } \right)\]
Given, \[{\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha \]
\[ \Rightarrow {\cos ^{ - 1}}\left( {x.{y \over 2} + \sqrt {1 - {x^2}} .\sqrt {1 - {{{y^2}} \over 4}} } \right) = \alpha \]
\[ \Rightarrow {{xy} \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{y{}^2} \over 4}} = \cos \,x\]
\[ \Rightarrow {\left( {\cos x - {{xy} \over 2}} \right)^2} = \left( {1 - {x^2}} \right)\left( {1 - {{{y^2}} \over 4}} \right)\]
\[ \Rightarrow {\cos ^2} + {{{x^2}{y^2}} \over 4} - 2.\cos x.{{xy} \over 2}\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - {x^2} - {{{y^2}} \over 4} + {{{x^2}{y^2}} \over 4}\]
\[ \Rightarrow {x^2} + {{{y^2}} \over 4} - xy\,\cos x = 1 - {\cos ^2}x\]
\[ \Rightarrow {x^2} + {{{y^2}} \over 4} - xy\cos x = {\sin ^2}x\]
\[ \Rightarrow 4{x^2} + y{}^2 - 4xy\cos x = 4{\sin ^2}x\]