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

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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