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

Question

Is this trigonometric identity provable?

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

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

Answer 1

$\cot\theta\sec\theta$

$=\frac{\cos\theta}{\sin\theta}\times\frac{1}{\cos\theta}$

$=\frac{1}{\sin\theta}$

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

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