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.
 A: Assuming the First Pythagorean Trigonometric Identity,
$$\sin^2\theta + \cos^2\theta = 1$$
Dividing by $\cos^2\theta$,
$$\Rightarrow \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}$$ 
$$\Rightarrow \left(\frac{\sin\theta}{\cos\theta}\right)^2 + \left(\frac{\cos\theta}{\cos\theta}\right)^2 = \left(\frac{1}{\cos\theta}\right)^2$$ 
Since $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ and $\sec\theta = \dfrac{1}{\cos\theta}$,
$$\Rightarrow \tan^2\theta + 1 = \sec^2\theta $$
Hence Proved.
A: 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,
$$
\begin{align*}
1+\tan^2 \theta &= 1+\left( \frac{y}{x} \right)^2 \\
&=1+\frac{y^2}{x^2} \\
&=\frac{x^2+y^2}{x^2} \\
&=\frac{r^2}{x^2} \\
&= \left(\frac{r}{x}\right)^2 \\
&=\sec^2 \theta.
\end{align*}
$$
A: $\cos(x-x)=\cos^2x+\sin^2x=1$ then divide by $\cos^2x$ to get the result above. I've assumed the one of the trigonometric results. 
A: Using the facts that $\frac d{d\theta}\tan\theta=\sec^2\theta$, $\frac d{d\theta} \sec\theta=\tan\theta\sec\theta$, $\tan 0 =0$, and $\sec 0=1$:
\begin{align*}
\frac d{d\theta} (1+\tan^2\theta)&=2\tan\theta\sec^2\theta\\
\frac d{d\theta} \sec^2\theta&=2\sec\theta(\tan\theta\sec\theta)\\
&=2\tan\theta\sec^2\theta.
\end{align*}
Thus $(1+\tan^2\theta)-\sec^2\theta$ is a constant. Since it is $0$ at $\theta=0$, it is zero everywhere.
