2
$\begingroup$

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.

$\endgroup$

4 Answers 4

7
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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).

A right triangle

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,

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

$\endgroup$
4
$\begingroup$

$\cos(x-x)=\cos^2x+\sin^2x=1$ then divide by $\cos^2x$ to get the result above. I've assumed the one of the trigonometric results.

$\endgroup$
2
  • 2
    $\begingroup$ That's a neat way to show Pythagoras' theorem! $\endgroup$ Dec 15, 2013 at 1:53
  • $\begingroup$ @user18921 thanks $\endgroup$
    – WhizKid
    Dec 15, 2013 at 12:08
2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .