Prove $1 + \tan^2\theta = \sec^2\theta$ Prove the following trigonometric identity: $$1 + \tan^2\theta = \sec^2\theta$$
I'm curious to know of the different ways of proving this depending on different characterizations of tangent and secant.

Here is an alternative using exponential forms:
$$
\begin{align*}
1+\tan^2 \theta &=1+\left( \frac{e^{i \theta}-e^{-i \theta}}{i\left( e^{i \theta}+e^{-i\theta} \right)} \right)^2 \\
&=1-\frac{\left( e^{i \theta}-e^{-i \theta} \right)^2}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\
&=\frac{\left( e^{i \theta}+e^{-i\theta} \right)^2-\left( e^{i \theta}-e^{-i \theta} \right)^2}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\
&=\frac{e^{2i\theta}+2+e^{-2i\theta}-e^{2i\theta}+2-e^{-2i\theta}}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\
&=\frac{4}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\
&=\left( \frac{2}{ e^{i \theta}+e^{-i\theta}} \right)^2 \\
&= \sec^2 \theta.
\end{align*}
$$

Here is an entirely different approach that focuses on the geometry of a right triangle.
Form a right triangle with angle $\theta$. Let $y$ be the side opposite $\theta$, $x$ be the side adjacent $\theta$, and label the hypotenuse $r$, where $r^2=x^2+y^2$ (by theorem of Pythagoras).

We can read trigonometric definitions right from the triangle as corresponding ratios of sides. Specifically, for angle $\theta$, $\tan \theta = \dfrac{y}{x}$, and $\sec \theta = \dfrac{r}{x}$. We can now write,
$$
\begin{align*}
1+\tan^2 \theta &= 1+\left( \frac{y}{x} \right)^2 \\
&=1+\frac{y^2}{x^2} \\
&=\frac{x^2+y^2}{x^2} \\
&=\frac{r^2}{x^2} \\
&= \left(\frac{r}{x}\right)^2 \\
&=\sec^2 \theta.
\end{align*}
$$
