Dec 7, 2017

See explanation.

#### Explanation:

Assuming you mean verify, not simplify, let's proceed.

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

When verifying an equation, it is good practice to only work one side of the equation and turn it into the other side. Let's turn the left side into the right side.

First, multiply each fraction by a clever form of $1$, noting that $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$.
$\frac{1}{1 - \cos \theta} \textcolor{g r e e n}{\times \frac{1 + \cos \theta}{1 + \cos \theta}} + \frac{1}{1 + \cos \theta} \textcolor{g r e e n}{\times \frac{1 - \cos \theta}{1 - \cos \theta}} = \frac{1 + \cos \theta}{1 - {\cos}^{2} \theta} + \frac{1 - \cos \theta}{1 - {\cos}^{2} \theta}$

Now combine like terms.

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

The Pythagorean identity states that ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$. This can be rearranged to ${\sin}^{2} \theta = 1 - {\cos}^{2} \theta$, so our expression is equal to $\frac{2}{\sin} ^ 2 \theta$.

Since $\csc \theta = \frac{1}{\sin} \theta$, $\frac{1}{\sin} ^ 2 \theta = {\csc}^{2} \theta$ so our expression is equal to $2 {\csc}^{2} \theta$, which is what we wanted to prove.