# How do you prove that (csc x - cot x)^2 = (1 - cos x)/(1+cosx) ?

Dec 12, 2015

Only modify the left side.

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

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

$= \frac{1 - 2 \cos x + {\cos}^{2} x}{{\sin}^{2} x}$

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

=(1-cosx)^2/((1+cosx)(1-cosx)

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