# How do you verify the identity (csctheta-cottheta)(csctheta+cottheta)=1?

Sep 9, 2016

We have: $\left(\csc \left(\theta\right) - \cot \left(\theta\right)\right) \left(\csc \left(\theta\right) + \cot \left(\theta\right)\right)$

Let's expand the parentheses:

= (csc(theta)) (csc(theta)) + (csc(theta)) (cot(theta)) + ( - cot(theta) (csc(theta)) + (- cot(theta)) (cot(theta))

$= {\csc}^{2} \left(\theta\right) + \csc \left(\theta\right) \cot \left(\theta\right) - \csc \left(\theta\right) \cot \left(\theta\right) - {\cot}^{2} \left(\theta\right)$

${\csc}^{2} \left(\theta\right) - {\cot}^{2} \left(\theta\right)$

Then, let's apply two standard trigonometric identities; $\csc \left(\theta\right) = \frac{1}{\sin \left(\theta\right)}$ and $\cot \left(\theta\right) = \frac{\cos \left(\theta\right)}{\sin \left(\theta\right)}$:

$= {\left(\frac{1}{\sin \left(\theta\right)}\right)}^{2} - {\left(\frac{\cos \left(\theta\right)}{\sin \left(\theta\right)}\right)}^{2}$

$= \frac{1}{{\sin}^{2} \left(\theta\right)} - \frac{{\cos}^{2} \left(\theta\right)}{{\sin}^{2} \left(\theta\right)}$

$= \frac{1 - {\cos}^{2} \left(\theta\right)}{{\sin}^{2} \left(\theta\right)}$

One of the Pythagorean identities is ${\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) = 1$.

We can rearrange this to get:

$\implies {\sin}^{2} \left(\theta\right) = 1 - {\cos}^{2} \left(\theta\right)$

Let's apply this rearranged identity to our proof:

$= \frac{{\sin}^{2} \left(\theta\right)}{{\sin}^{2} \left(\theta\right)}$

$= 1$ (Q.E.D.)

Sep 9, 2016

I changed all into $\sin$ and $\cos$:

#### Explanation:

have a look: 