Answer
$${A_{{R_1}}} > {A_{{R_2}}}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}{A_{{R_1}}}{\text{ the area of the region }}{R_1} \cr
& y = \tan x,{\text{ on the interval }}\left[ {0,\frac{\pi }{3}} \right],{\text{ then}} \cr
& {A_{{R_1}}} = \int_0^{\pi /3} {\tan x} dx \cr
& {\text{Integrate}} \cr
& {A_{{R_1}}} = \left[ {\ln \left| {\sec x} \right|} \right]_0^{\pi /3} \cr
& {A_{{R_1}}} = \ln \left| {\sec \left( {\frac{\pi }{3}} \right)} \right| - \ln \left| {\sec \left( 0 \right)} \right| \cr
& {A_{{R_1}}} = \ln \left| 2 \right| - \ln \left| 1 \right| \cr
& {A_{{R_1}}} = \ln 2 \cr
& \cr
& {\text{Let }}{A_{{R_2}}}{\text{ the area of the region }}{R_2} \cr
& y = \sec x,{\text{ on the interval }}\left[ {0,\frac{\pi }{6}} \right],{\text{ then}} \cr
& {A_{{R_2}}} = \int_0^{\pi /6} {\sec x} dx \cr
& {\text{Integrate}} \cr
& {A_{{R_2}}} = \left[ {\ln \left| {\sec x + \tan x} \right|} \right]_0^{\pi /6} \cr
& {A_{{R_2}}} = \ln \left| {\sec \left( {\frac{\pi }{6}} \right) + \tan \left( {\frac{\pi }{6}} \right)} \right| - \ln \left| {\sec \left( 0 \right) + \tan \left( 0 \right)} \right| \cr
& {A_{{R_2}}} = \ln \left| {\frac{{2\sqrt 3 }}{3} + \frac{{\sqrt 3 }}{3}} \right| - \ln \left| 1 \right| \cr
& {A_{{R_2}}} = \ln \sqrt 3 \cr
& \cr
& \ln 2 > \ln \sqrt 3 ,{\text{ Therefore }}{A_{{R_1}}} > {A_{{R_2}}} \cr} $$
