Official solution
\[\int\limits_0^{\pi /2} {{{\cot x} \over {\cot x + {\mathop{\rm cosec}\nolimits} \,x}}dx = } \int\limits_0^{\pi /2} {{{\cos x} \over {\cos x + 1}}dx} \]
\[ \Rightarrow \int\limits_0^{\pi /2} {{{2{{\cos }^2}{x \over 2} - 1} \over {2{{\cos }^2}{x \over 2}}}dx = } \int\limits_0^{\pi /2} {\left( {1 - {1 \over 2}{{\sec }^2}{x \over 2}} \right)dx} \]
\[ \Rightarrow \left( {x - \tan {x \over 2}} \right)_0^{\pi /2} = {\pi \over 2} - 1 = {1 \over 2}\left( {\pi - 2} \right)\]
mn = \[{1 \over 2}x - 2 = - 1\]