Official solution
\[f\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin \,x} \over {1 - \sin x}}} } \right)\]
\[ = {\tan ^{ - 1}}\left( {\sqrt {{{{{\left( {\sin {x \over 2} + \cos {x \over 2}} \right)}^2}} \over {{{\left( {\sin {x \over x} - \cos {x \over 2}} \right)}^2}}}} } \right)\]
\[ = {\tan ^{ - 1}}\left( {{{1 + \tan {x \over 2}} \over {1 - \tan {x \over 2}}}} \right)\]
\[ = {\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} + {x \over 2}} \right)} \right)\]
\[ \Rightarrow y = {\pi \over 4} + {x \over 2}\]
\[ \Rightarrow {{dy} \over {dx}} = {1 \over 2}\]
Slope of normal \[ = {{ - 1} \over {\left( {{{dy} \over {dx}}} \right)}} = - 2\]
At \[\left( {{\pi \over 6},{\pi \over 4} + {\pi \over {12}}} \right)\]
\[y - \left( {{\pi \over 4} + {\pi \over {12}}} \right) = - 2\left( {x - {\pi \over 6}} \right)\]
\[y - {{4\pi } \over {12}} = - 2x + {{2\pi } \over 6}\]
\[y - {\pi \over 3} = - 2x + {\pi \over 3}\]
\[y = - 2x + {{2\pi } \over 3}\]
This equation is satisfied only by the point
\[\left( {0,{{2\pi } \over 3}} \right)\]