From the pythagorean identity we have that
#sin^2x + cos^2x =1#
Dividing both sides by #cos^2x# we have
#tan^2x + 1 = 1/cos^2x#
Which means that, if we isolate the tangent we have
#tan^2x = 1/cos^2x - 1#
So for #x = arccos(2/3)/2# we have
#tan^2(arccos(2/3)/2) = 1/cos^2(arccos(2/3)/2) -1#
#cos^2(arccos(2/3)/2)# can be calculated using the half angle formula, so we know that
#cos^2(arccos(2/3)/2) = (1+cos(arccos(2/3)))/2#
Since #cos(arccos(x)) = x# we can say that
#cos^2(arccos(2/3)/2) = (1+2/3)/2 = (5/3)/2 = 5/3*1/2 = 5/6#
Putting that back on the formula
#tan^2(arccos(2/3)/2) = 1/(5/6) -1 = 6/5 - 1 = (6-5)/5 = 1/5#
Taking the root
#tan(arccos(2/3)/2) = +-sqrt(1/5)#
Knowing that during the range of the arccosine, the tangent is only negative if the cosine's also negative we have that
#tan(arccos(2/3)/2) = 1/sqrt(5) = sqrt(5)/5#
