# How do you identify 1/(csctheta+1)-1/(csctheta-1)?

$- 2 {\tan}^{2} \theta$

#### Explanation:

I'm going to assume this needs to be simplified:

$\frac{1}{\csc \theta + 1} - \frac{1}{\csc \theta - 1}$

$\frac{1}{\frac{1}{\sin} \theta + 1} - \frac{1}{\frac{1}{\sin} \theta - 1}$

$\frac{1}{\frac{1}{\sin} \theta + \sin \frac{\theta}{\sin} \theta} - \frac{1}{\frac{1}{\sin} \theta - \sin \frac{\theta}{\sin} \theta}$

$\frac{1}{\frac{1 + \sin \theta}{\sin} \theta} - \frac{1}{\frac{1 - \sin \theta}{\sin} \theta}$

$\sin \frac{\theta}{1 + \sin \theta} - \sin \frac{\theta}{1 - \sin \theta}$

$\sin \frac{\theta}{1 + \sin \theta} \left(\frac{1 - \sin \theta}{1 - \sin \theta}\right) - \sin \frac{\theta}{1 - \sin \theta} \left(\frac{1 + \sin \theta}{1 + \sin \theta}\right)$

$\frac{\sin \theta \left(1 - \sin \theta\right)}{\left(1 + \sin \theta\right) \left(1 - \sin \theta\right)} - \frac{\sin \theta \left(1 + \sin \theta\right)}{\left(1 - \sin \theta\right) \left(1 + \sin \theta\right)}$

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

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

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

Recall that ${\sin}^{2} \theta + {\cos}^{2} \theta = 1 \implies 1 - {\sin}^{2} \theta = {\cos}^{2} \theta$

$\frac{- 2 {\sin}^{2} \theta}{\cos} ^ 2 \theta = - 2 {\tan}^{2} \theta$