So you want to express #tan(x)# in terms of #sec(x)#.

Well, first of all, let's write #tan(x)# in a form that we are more familiar with:

#tan(x)=sin(x)/cos(x)#

We also note that #sec(x)=1/cos(x)#

So, already, we see that

#tan(x)=sin(x)*sec(x)#

Now we only need to express #sin(x)# in terms of #sec(x)#.

We know that

#sin^2(x) + cos^2(x) = 1#

so

#sin^2(x) = 1 - cos^2(x)#

which means that

#sin(x) = pm sqrt(1-cos^2(x))#

which means that

#sin(x) = pm sqrt(1- 1/sec^2(x))#

We finally have:

#tan(x) = pm sqrt(1- 1/sec^2(x)) * sec(x)#

We could also absorb the #sec(x)# in the square-root, giving us

#tan(x) = pm sqrt(sec^2(x) - 1)#

It is more common to write it as follows:

#tan^2(x) + 1 = sec^2(x)#

We have expressed #tan(x)# in terms of #sec(x)#.

Q.E.D.