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: $\dfrac{1}{\tan\theta \cos\theta}$
 A: $\cot\theta\sec\theta$
$=\frac{\cos\theta}{\sin\theta}\times\frac{1}{\cos\theta}$
$=\frac{1}{\sin\theta}$
A: Hint: Do you remember the relationship between $\tan{\theta}$, $\cos{\theta}$, and $\sin{\theta}$?
A: 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.
A: secant is $ \frac{1}{cos(\theta)}$.  so multiply both sides by $sin (\theta)$.  You get $\frac{sin(\theta)}{cos(\theta)} = tan(\theta)$.
now clearly $cot(\theta)*tan(\theta) = 1$.
A: $$\cot \theta \sec\theta= \frac{adjacent}{opposite} \frac{hypotenuse}{adjacent} = \frac{hypotenuse}{opposite} = 1/ \sin\theta$$
