# Prove $1 + \cot^2\theta = \csc^2\theta$

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.

Consider the diagram below. 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*}

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 $$f\left(k\pi+\frac{\pi}{2}\right)=1$$

• Nice work, and great "tie-in" with the linked post! +1 – amWhy Dec 16 '13 at 14:47

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.

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