Answer
$$K = \frac{4}{{5\sqrt 5 }},{\text{ }}r = \frac{{5\sqrt 5 }}{4}$$
Work Step by Step
$$\eqalign{
& y = \tan x,{\text{ }}x = \frac{\pi }{4} \cr
& {\text{Calculate the curvature, use }}K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Find }}y'\left( x \right){\text{ and }}y''\left( x \right){\text{}} \cr
& y'\left( x \right) = \frac{d}{{dx}}\left[ {\tan x} \right] \cr
& y'\left( x \right) = {\sec ^2}x \cr
& {\text{Evaluate at }}x = \frac{\pi }{4} \cr
& y'\left( {\frac{\pi }{4}} \right) = {\sec ^2}\left( {\frac{\pi }{4}} \right) = 2 \cr
& y''\left( x \right) = \frac{d}{{dx}}\left[ {{{\sec }^2}x} \right] \cr
& y''\left( x \right) = 2{\sec ^2}x\tan x \cr
& {\text{Evaluate at }}x = \frac{\pi }{4} \cr
& y''\left( {\frac{\pi }{4}} \right) = 2{\sec ^2}\left( {\frac{\pi }{4}} \right)\tan \left( {\frac{\pi }{4}} \right) \cr
& y''\left( {\frac{\pi }{4}} \right) = 2\left( 2 \right)\left( 1 \right) = 4 \cr
& \underbrace {K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}}}_ \Downarrow \cr
& {\text{ at }}x = \frac{\pi }{4} \cr
& K = \frac{{\left| 4 \right|}}{{{{\left( {1 + {{\left[ 2 \right]}^2}} \right)}^{3/2}}}} \cr
& K = \frac{4}{{5\sqrt 5 }} \cr
& {\text{The radius of curvature is }}r = \frac{1}{K} \cr
& r = \frac{{5\sqrt 5 }}{4} \cr} $$
