Proof of trigonometric identity $\cot \theta \sec\theta= 1/ \sin\theta$

Is this trigonometric identity provable?

$$\color{red}{}\;\color{navy}{\cot \theta \sec \theta = \dfrac 1 {\sin \theta}}$$

I can't seem to get passed: $\dfrac{1}{\tan\theta \cos\theta}$

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What is the definition of $\ \tan \theta \$ ? What happens when you mulitply it by $\ \cos \theta \$ ? –  RecklessReckoner Aug 7 at 1:05
You might wish to note that (most simple) statements are true or false. In this case it is either provable or an axiom. You might wish to think about the problem as "how should I prove," rather than "is it possible to prove." –  Thoth19 Aug 7 at 2:49

\begin{align}\cot\theta\sec\theta &= \frac{\cos\theta}{\sin\theta}\times\frac{1}{\cos\theta}\\ &=\frac{1}{\sin\theta}\end{align}

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Hint: Do you remember the relationship between $\tan{\theta}$, $\cos{\theta}$, and $\sin{\theta}$?

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secant is $\frac{1}{cos(\theta)}$. so multiply both sides by $sin (\theta)$. You get $\frac{sin(\theta)}{cos(\theta)} = tan(\theta)$.
now clearly $cot(\theta)*tan(\theta) = 1$.

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I don't know how to divide out by trig functions so I convert everyting to $e^{\rm something}$.

We know that: $$\sin \theta = \frac{i e^{- i \theta} - i e^{ i \theta}}{2}$$

$$\cos \theta = \frac{e^{- i \theta} + e^{ i \theta}}{2}$$

and $\cot \theta = \frac{1}{\tan\theta} = \frac{1}{\frac{\sin\theta}{\cos\theta}}$ and $\sec \theta = \frac{1}{\cos\theta}$ as you already know.

So know we can do everything in terms of arithmetic on addition, multiplication, and exponentials instead of those pesky inscrutable trigonometrics.

$$\cot\theta \cdot\sec\theta = \frac{{\frac{e^{- i \theta} + e^{ i \theta}}{2}}}{\frac{i e^{-i\theta} - i e^{i\theta}}{2}} \cdot \frac{1}{\frac{e^{- i \theta} + e^{ i \theta}}{2}} = \frac{1}{\frac{i e^{- i \theta} - i e^{ i \theta}}{2}} = \frac{1}{\sin\theta}$$

This method will work for any trigonometric identity. You may have to use some arithmetic tricks, but no weird trig functions.

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$$\cot \theta \sec\theta= \frac{adjacent}{opposite} \frac{hypotenuse}{adjacent} = \frac{hypotenuse}{opposite} = 1/ \sin\theta$$

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