2
$\begingroup$

Prove the following identity: $$1 + \cot^2\theta = \csc^2\theta$$

This question is asked because I am curious to know the different ways of proving this identity depending on different characterizations of cotangent and cosecant.

$\endgroup$
2
$\begingroup$

Consider the diagram below.Pythagorean_identity_for_cotangent_cosecant

The terminal side of an angle $\theta$ in standard position intersects the unit circle at the point $(\cos\theta, \sin\theta)$. If

$$\theta \neq \frac{\pi}{2} + n\pi, n \in \mathbb{Z}$$

draw a segment perpendicular to the terminal side of the angle. Since two angles complementary to the same angle are congruent, the angle formed by the perpendicular to the terminal side of the angle $\theta$ in standard position and the $y$-axis has measure $\theta$. Since the leg opposite angle $\theta$ in this triangle has length $1$, the hypotenuse has length $|\csc\theta|$ and the leg adjacent to angle $\theta$ has length $|\cot\theta|$. By the Pythagorean Theorem,

\begin{align*} 1 + |\cot\theta|^2 & = |\csc\theta|^2\\ 1 + \cot^2\theta & = \csc^2\theta \end{align*}

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

Here we repeat an idea used in the question Prove $\sin^2\theta+\cos^2\theta=1$ but it's slightly different since the functions $\cot$ and $\csc$ aren't defined on $\mathbb R$.

Let $$f(\theta)=\csc^2\theta-\cot^2\theta$$ then $f$ is defined on $\mathbb R\setminus\{k\pi,\; k\in\mathbb Z\}$ and we verify that $f'(\theta)=0$ so $f$ is constant in every interval $(k\pi,(k+1)\pi)$ and we conclude the result from the equality $$f\left(k\pi+\frac{\pi}{2}\right)=1$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Nice work, and great "tie-in" with the linked post! +1 $\endgroup$ – amWhy Dec 16 '13 at 14:47
3
$\begingroup$

You could also start from left to right.

$$ \begin{align*} 1 + \cot^2 \theta & = 1 + \frac{\cos^2 \theta}{\sin^2 \theta} \\ & = \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta}\\ & = \frac{1}{\sin^2 \theta}\\ & = \csc^2 \theta~. \end{align*} $$

Just go backwards if you want to prove from right to left.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

Assuming the First Pythagorean Trigonometric Identity,

$$\sin^2\theta + \cos^2\theta = 1$$ Dividing by $\sin^2\theta$, $$\Rightarrow \frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta}$$ $$\Rightarrow \left(\frac{\sin\theta}{\sin\theta}\right)^2 + \left(\frac{\cos\theta}{\sin\theta}\right)^2 = \left(\frac{1}{\sin\theta}\right)^2$$

Since $\cot\theta = \large \frac{1}{\tan\theta} = \large\frac{\cos\theta}{\sin\theta}$ and $\csc\theta = \large\frac{1}{\sin\theta}$, $$\Rightarrow 1 + \cot^2\theta = \csc^2\theta .$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.