How do you prove cos2t=(1-tan^2 t)/(1+tan^2 t)?

Remembering that ${\sec}^{2} \left(t\right) - 1 = {\tan}^{2} \left(t\right)$ and the identity
$\cos \left(2 t\right) = {\cos}^{2} \left(t\right) - {\sin}^{2} \left(t\right)$ as well as each of the reciprocal identities, namely
$\tan t = \sin \frac{t}{\cos} t$ and $\sec t = \frac{1}{\cos} t$, we can start off by simplifying our equation on the right-hand side.
$\cos \left(2 t\right) = \frac{1 - {\sin}^{2} \frac{t}{{\cos}^{2} \left(t\right)}}{\frac{1}{\cos} ^ 2 \left(t\right)} = \frac{\frac{1}{1} \cdot {\cos}^{2} \frac{t}{1}}{\cancel{\frac{1}{\cos} ^ 2 \left(t\right) \cdot {\cos}^{2} \frac{t}{1}}} - \frac{{\sin}^{2} \frac{t}{\cancel{{\cos}^{2} \left(t\right)}} \cdot \cancel{{\cos}^{2} \frac{t}{1}}}{\frac{1}{\cancel{{\cos}^{2} \left(t\right)}} \cdot \cancel{{\cos}^{2} \frac{t}{1}}}$
$= {\cos}^{2} \left(t\right) - {\sin}^{2} \left(t\right)$