# How do you prove the identity (csc theta - 1) / cot theta = cot theta / (csc theta + 1)?

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