Prove the Identity: (csc α - cot α)^2 = (1-cos α)/(1+cos α)

$(csc \alpha -cot \alpha {)}^2 = \frac{1-cos \alpha }{1+cos \alpha }$ 

= ${\csc }^2\alpha - 2cot\alpha csc\alpha + co{t}^2\alpha$

Note:
$\csc \alpha =\frac{1}{\sin \alpha }$
and...
$\cot \alpha =\frac{\cos \alpha }{\sin \alpha }$

= ${\csc }^2\alpha - 2\left(\frac{\cos \alpha }{\sin \alpha }\right)\left(\frac{1}{\sin \alpha }\right) + \frac{co{s}^2\alpha }{{\sin }^2\alpha }$  

= $\frac{1}{{\sin }^2\alpha } - \frac{2cos\alpha }{{\sin }^2\alpha }+ \frac{co{s}^2\alpha }{{\sin }^2\alpha }$ 

= $\frac{1 - 2cos\alpha + co{s}^2\alpha }{{\sin }^2\alpha }$ 
 

Note: 
${\sin }^2\alpha = 1-co{s}^2\alpha$ 
and...
$1-2cos\alpha +co{s}^2\alpha =(1-cos\alpha )(1+cos\alpha )$

= $\frac{(1-cos\alpha )(1+cos\alpha )}{1-co{s}^2\alpha }$   

Note: 
$1-co{s}^2\alpha = (1-cos\alpha )(1+cos\alpha )$

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

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