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