# How do you verify (1+cotalpha)^2-2cotalpha=1/((1-cosalpha)(1+cosalpha))?

Dec 6, 2016

Start by transforming all the terms into sine and cosine using the identity color(red)(cot theta = 1/tantheta = 1/(sintheta/costheta) = costheta/sintheta

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

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

The $\frac{2 \cos \alpha}{\sin} \alpha$'s cancel each other out

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

We use the identity color(red)(sin^2beta + cos^2beta =1-> sin^2beta = 1- cos^2beta at this point in the process.

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

$\frac{1}{\sin} ^ 2 \alpha = \frac{1}{\sin} ^ 2 \alpha$

$L H S = R H S$

Identity Proved!

Hopefully this helps!

Dec 6, 2016

$L H S = {\left(1 + \cot \alpha\right)}^{2} - 2 \cot \alpha$

$= {1}^{2} + {\cot}^{2} \alpha + 2 \cot \alpha - 2 \cot \alpha$

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

$= \frac{1}{\sin} ^ 2 \alpha$

$= \frac{1}{1 - {\cos}^{2} \alpha}$

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

Proved