Question #81894

1 Answer
Pythagoras
Oct 7, 2017

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

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.

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