From the #color(red)(cos^-1(1/2) # we have the side #color(red)(adjacent# to #color(red)(color(red)(angle theta=1# and the side #color(red)(hypote n use = 2# . So this is a #color(red)(30^@-60^@-90^@# triangle and #color(red)(theta = 60^@#therefore the #color(red)(opposite# to #color(red)(angle theta = sqrt3#.
Note that from the restrictions for the range of inverse circular functions #color(red)(cos^-1 x# is restricted to quadrants 1 and 2 and since the argument is positive #color(red)(1/2)# our answer will be in quadrant 1 only.
Hence,
#color(blue)(tan(cos^-1(1/2))=tan theta=(opposite)/(adjacent)=sqrt3/1=sqrt3#