# How do you verify (1 - cosx) / (1 + cosx) = (cotx - cscx)²?

Nov 1, 2015

Have a look:

#### Explanation:

Consider that:
$\cot \left(x\right) = \cos \frac{x}{\sin} \left(x\right)$
and:
$\csc \left(x\right) = \frac{1}{\sin} \left(x\right)$
$\frac{1 - \cos \left(x\right)}{1 + \cos \left(x\right)} = {\left(\cos \frac{x}{\sin} \left(x\right) - \frac{1}{\sin} \left(x\right)\right)}^{2}$
$\frac{1 - \cos \left(x\right)}{1 + \cos \left(x\right)} = {\left(\frac{\cos \left(x\right) - 1}{\sin} \left(x\right)\right)}^{2}$
$\frac{1 - \cos \left(x\right)}{1 + \cos \left(x\right)} = {\left(\cos \left(x\right) - 1\right)}^{2} / {\sin}^{2} \left(x\right)$
but simplyfying and from:
$1 = {\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right)$
${\sin}^{2} \left(x\right) = 1 - {\cos}^{2} \left(x\right)$
$\frac{\cancel{\left(1 - \cos \left(x\right)\right)}}{1 + \cos \left(x\right)} = {\left(\cos \left(x\right) - 1\right)}^{\cancel{2}} / \left(1 - {\cos}^{2} \left(x\right)\right)$
also:
$\left(1 - {\cos}^{2} \left(x\right)\right) = \left(1 + \cos \left(x\right)\right) \left(1 - \cos \left(x\right)\right)$
$\frac{1}{1 + \cos \left(x\right)} = \frac{\cos \left(x\right) - 1}{\left(1 + \cos \left(x\right)\right) \left(1 - \cos \left(x\right)\right)}$
and finally:
$\frac{1}{\cancel{\left(1 + \cos \left(x\right)\right)}} = \frac{\cancel{\left(\cos \left(x\right) - 1\right)}}{\cancel{\left(1 + \cos \left(x\right)\right)} \cancel{\left(1 - \cos \left(x\right)\right)}}$
or:
$1 = 1$