# How do you prove cot(theta)sec(theta)=csc(theta)?

May 30, 2016

Applying the definition of $\cot , \sec , \csc$ and $\tan$.

#### Explanation:

Recall the definition of these functions:

$\cot \left(\theta\right) = \frac{1}{\tan} \left(\theta\right)$
$\sec \left(\theta\right) = \frac{1}{\cos} \left(\theta\right)$
$\csc \left(\theta\right) = \frac{1}{\sin} \left(\theta\right)$

Then we want to prove

$\cot \left(\theta\right) \sec \left(\theta\right) = \csc \left(\theta\right)$

that is equivalent to

$\frac{1}{\tan} \left(\theta\right) \frac{1}{\cos} \left(\theta\right) = \frac{1}{\sin} \left(\theta\right)$

We recall that $\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right)$, consequently
$\frac{1}{\tan} \left(\theta\right) = \cos \frac{\theta}{\sin} \left(\theta\right)$.
I substitute in the previous equation

$\frac{1}{\tan} \left(\theta\right) \frac{1}{\cos} \left(\theta\right) = \frac{1}{\sin} \left(\theta\right)$

$\cos \frac{\theta}{\sin} \left(\theta\right) \frac{1}{\cos} \left(\theta\right) = \frac{1}{\sin} \left(\theta\right)$

$\frac{1}{\sin} \left(\theta\right) = \frac{1}{\sin} \left(\theta\right)$.