Question #81894

Oct 7, 2017

${\tan}^{2} \left(x\right) + 1 = {\sec}^{2} \left(x\right)$
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Explanation:

So you want to express $\tan \left(x\right)$ in terms of $\sec \left(x\right)$.

Well, first of all, let's write $\tan \left(x\right)$ in a form that we are more familiar with:
$\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$

We also note that $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$

$\tan \left(x\right) = \sin \left(x\right) \cdot \sec \left(x\right)$

Now we only need to express $\sin \left(x\right)$ in terms of $\sec \left(x\right)$.
We know that
${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$
so
${\sin}^{2} \left(x\right) = 1 - {\cos}^{2} \left(x\right)$
which means that
$\sin \left(x\right) = \pm \sqrt{1 - {\cos}^{2} \left(x\right)}$
which means that
$\sin \left(x\right) = \pm \sqrt{1 - \frac{1}{\sec} ^ 2 \left(x\right)}$

We finally have:
$\tan \left(x\right) = \pm \sqrt{1 - \frac{1}{\sec} ^ 2 \left(x\right)} \cdot \sec \left(x\right)$

We could also absorb the $\sec \left(x\right)$ in the square-root, giving us
$\tan \left(x\right) = \pm \sqrt{{\sec}^{2} \left(x\right) - 1}$

It is more common to write it as follows:
${\tan}^{2} \left(x\right) + 1 = {\sec}^{2} \left(x\right)$

We have expressed $\tan \left(x\right)$ in terms of $\sec \left(x\right)$.
Q.E.D.