# How do you verify the identity (csc x - cot x)^2 = (1 - cos x)/(1+cosx)?

Mar 6, 2018

See below

#### Explanation:

${\left(\csc - \cot x\right)}^{2} = \frac{1 - \cos x}{1 + \cos x}$
$\left({\csc}^{2} x - 2 \cot x \csc x + {\cot}^{2} x\right) = \frac{1 - \cos x}{1 + \cos x}$
$\left(\frac{1}{\sin} ^ 2 x - \frac{2 \cos x}{\sin} ^ 2 x + {\cos}^{2} \frac{x}{\sin} ^ 2 x\right) = \frac{1 - \cos x}{1 + \cos x}$

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

$\left(\frac{\left(1 - \cos x\right) \left(1 - \cos x\right)}{1 - {\cos}^{2} x}\right) = \frac{1 - \cos x}{1 + \cos x}$

$\left(\frac{\cancel{1 - \cos x} \left(1 - \cos x\right)}{\cancel{1 - \cos x} \left(1 + \cos x\right)}\right) = \frac{1 - \cos x}{1 + \cos x}$

$\frac{1 - \cos x}{1 + \cos x} = \frac{1 - \cos x}{1 + \cos x}$