Question

Question #81894

Answer 1

`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.