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#