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

May 17, 2015

The $-$ sign in the question should be a $+$.

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

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

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

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

$= \frac{2}{{\sin}^{2} \theta}$

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

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

$= 2 {\csc}^{2} \theta$

If you try the same thing with $\frac{1}{1 - \cos \theta} - \frac{1}{1 + \cos \theta}$ you will get the result $2 \cos \theta {\csc}^{2} \theta$.