Official solution
\[{{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}\]
\[{{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \over {a + b}} = {\pi \over {2(a + b)}}\]
Now, \[{{{{\tan }^{ - 1}}y} \over c} = {\pi \over {2(a + b)}}\]
\[2{\tan ^{ - 1}}y = {{\pi c} \over {a + b}}\]
\[ \Rightarrow \cos \left( {{{\pi c} \over {a + b}}} \right) = \cos (2{\tan ^{ - 1}}y) = {{1 - {y^2}} \over {1 + {y^2}}}\]