# How do you verify the identity (csctheta-cottheta)^2=(1-costheta)/(1+costheta)?

Aug 21, 2016

the identity is verified

#### Explanation:

Since $\csc \theta = \frac{1}{\sin} \theta \mathmr{and} \cot \theta = \cos \frac{\theta}{\sin \theta}$,

you can substitute and have:

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

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

Since ${\sin}^{2} \theta = 1 - {\cos}^{2} \theta$, the expression becomes:

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

and the identity is verified