What is tan^2theta in terms of non-exponential trigonometric functions?

${\tan}^{2} \left(\theta\right) = \frac{1 - \cos \left(2 \theta\right)}{1 + \cos \left(2 \theta\right)}$
You first need to remember that $\cos \left(2 \theta\right) = 2 {\cos}^{2} \left(\theta\right) - 1 = 1 - 2 {\sin}^{2} \left(\theta\right)$. Those equalities give you a "linear" formula for ${\cos}^{2} \left(\theta\right)$ and ${\sin}^{2} \left(\theta\right)$.
We now know that ${\cos}^{2} \left(\theta\right) = \frac{1 + \cos \left(2 \theta\right)}{2}$ and ${\sin}^{2} \left(\theta\right) = \frac{1 - \cos \left(2 \theta\right)}{2}$ because $\cos \left(2 \theta\right) = 2 {\cos}^{2} \left(\theta\right) - 1 \iff 2 {\cos}^{2} \left(\theta\right) = 1 + \cos \left(2 \theta\right) \iff {\cos}^{2} \left(\theta\right) = \frac{1 + \cos \left(2 \theta\right)}{2}$. Same for ${\sin}^{2} \left(\theta\right)$.
${\tan}^{2} \left(\theta\right) = {\sin}^{2} \frac{\theta}{\cos} ^ 2 \left(\theta\right) = \frac{1 - \cos \left(2 \theta\right)}{2} \cdot \frac{2}{1 + \cos \left(2 \theta\right)} = \frac{1 - \cos \left(2 \theta\right)}{1 + \cos \left(2 \theta\right)}$