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 = $${1 \over 2}$$, $${\left( {{{df} \over {dg}}} \right)_{x = {1 \over 2}}}$$ = $${{\sqrt 3 } \over {10}}$$