Official solution
Let f = \[{\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)\]
Put x = tan \[\theta \] \[ \Rightarrow \] \[\theta \] = tan--1 x
f = \[{\tan ^{ - 1}}\left( {{{\sec \theta - 1} \over {\tan \theta }}} \right)\]
\[ \Rightarrow \] f = \[{\tan ^{ - 1}}\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right)\] = \[{\theta \over 2}\]
\[ \Rightarrow \] f = \[{{{{\tan }^{ - 1}}x} \over 2}\]
\[ \therefore \] \[{{df} \over {dx}}\] = \[{1 \over {2\left( {1 + {x^2}} \right)}}\] ....(1)
Let g = \[{\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right)\]
Put x = sin \[\theta \] \[ \Rightarrow \] \[\theta \] = sin--1
x
\[ \Rightarrow \] g = \[{\tan ^{ - 1}}\left( {{{2\sin \theta \cos \theta } \over {1 - 2{{\sin }^2}\theta }}} \right)\]
\[ \Rightarrow \] g = tan--1
(tan 2\[\theta \]) = 2\[\theta \]
\[ \Rightarrow \] g = 2sin-1 x
\[ \Rightarrow \] \[{{dg} \over {dx}}\] = \[{2 \over {\sqrt {1 - {x^2}} }}\] ...(2)
Using (i) and (ii),
\[ \therefore \] \[{{df} \over {dg}}\] = \[{1 \over {2\left( {1 + {x^2}} \right)}}{{\sqrt {1 - {x^2}} } \over 2}\]
At x = \[{1 \over 2}\], \[{\left( {{{df} \over {dg}}} \right)_{x = {1 \over 2}}}\] = \[{{\sqrt 3 } \over {10}}\]