[Math] Prove $1 + \tan^2\theta = \sec^2\theta$

Question

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.

Answer 1

Here is an alternative using exponential forms:

$$ \begin{align*} 1+\tan^2 \theta &=1+\left( \frac{e^{i \theta}-e^{-i \theta}}{i\left( e^{i \theta}+e^{-i\theta} \right)} \right)^2 \\ &=1-\frac{\left( e^{i \theta}-e^{-i \theta} \right)^2}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\ &=\frac{\left( e^{i \theta}+e^{-i\theta} \right)^2-\left( e^{i \theta}-e^{-i \theta} \right)^2}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\ &=\frac{e^{2i\theta}+2+e^{-2i\theta}-e^{2i\theta}+2-e^{-2i\theta}}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\ &=\frac{4}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\ &=\left( \frac{2}{ e^{i \theta}+e^{-i\theta}} \right)^2 \\ &= \sec^2 \theta. \end{align*} $$

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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,

$$ \begin{align*} 1+\tan^2 \theta &= 1+\left( \frac{y}{x} \right)^2 \\ &=1+\frac{y^2}{x^2} \\ &=\frac{x^2+y^2}{x^2} \\ &=\frac{r^2}{x^2} \\ &= \left(\frac{r}{x}\right)^2 \\ &=\sec^2 \theta. \end{align*} $$