![(\frac{1}{1-cos(x)})-(\frac{cos(x)}{1+cos(x)})](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B1-cos%28x%29%7D%29-%28%5Cfrac%7Bcos%28x%29%7D%7B1%2Bcos%28x%29%7D%29)
cscx is 1/sinx
maybe they want us to use pythagorean identity
![cos^2(x)+sin^2(x)=1](https://tex.z-dn.net/?f=cos%5E2%28x%29%2Bsin%5E2%28x%29%3D1)
I notice we have 1-cos(x) and 1+cos(x), if we multiply them, we get ![1-cos^2(x)](https://tex.z-dn.net/?f=1-cos%5E2%28x%29)
and if we look at the pythagorean identity and minus cos^2(x) from both sides, we get
![sin^2(x)=1-cos^2(x)](https://tex.z-dn.net/?f=sin%5E2%28x%29%3D1-cos%5E2%28x%29)
since
, ![csc^2(x)=\frac{1}{sin^2(x)}=\frac{1}{1-cos^2(x)}](https://tex.z-dn.net/?f=csc%5E2%28x%29%3D%5Cfrac%7B1%7D%7Bsin%5E2%28x%29%7D%3D%5Cfrac%7B1%7D%7B1-cos%5E2%28x%29%7D)
(recall that (a-b)(a+b)=a²-b²)
match the denomenators of the original fraction
multiply first fraction by
and the 2nd by ![\frac{1-cos(x)}{1-cos(x)}](https://tex.z-dn.net/?f=%5Cfrac%7B1-cos%28x%29%7D%7B1-cos%28x%29%7D)
![(\frac{1+cos(x)}{1-cos^2(x)})-(\frac{cos(x)(1-cos(x))}{1-cos^2(x)})=](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%2Bcos%28x%29%7D%7B1-cos%5E2%28x%29%7D%29-%28%5Cfrac%7Bcos%28x%29%281-cos%28x%29%29%7D%7B1-cos%5E2%28x%29%7D%29%3D)
![((csc^2(x))(1+cos(x)))-((csc^2(x))(cos(x)-cos^2(x)))=](https://tex.z-dn.net/?f=%28%28csc%5E2%28x%29%29%281%2Bcos%28x%29%29%29-%28%28csc%5E2%28x%29%29%28cos%28x%29-cos%5E2%28x%29%29%29%3D)
![csc^2(x)+csc^2(x)cos(x)-(csc^2(x)cos(x)-csc^2(x)cos^2(x)=](https://tex.z-dn.net/?f=csc%5E2%28x%29%2Bcsc%5E2%28x%29cos%28x%29-%28csc%5E2%28x%29cos%28x%29-csc%5E2%28x%29cos%5E2%28x%29%3D)
![csc^2(x)+csc^2(x)cos(x)-csc^2(x)cos(x)+csc^2(x)cos^2(x)=](https://tex.z-dn.net/?f=csc%5E2%28x%29%2Bcsc%5E2%28x%29cos%28x%29-csc%5E2%28x%29cos%28x%29%2Bcsc%5E2%28x%29cos%5E2%28x%29%3D)
![csc^2(x)+csc^2(x)cos^2(x)=](https://tex.z-dn.net/?f=csc%5E2%28x%29%2Bcsc%5E2%28x%29cos%5E2%28x%29%3D)
, hmm, to get ride of those cos(x)
look to the pythagorean identity again
, force one side into form 1+cos^2(x)
, recall that since csc(x)=1/sin(x), sin(x)=1/csc(x) and sin^2(x)=1/(csc^2(x))
![1+cos^2(x)=2-\frac{1}{csc^2(x)}](https://tex.z-dn.net/?f=1%2Bcos%5E2%28x%29%3D2-%5Cfrac%7B1%7D%7Bcsc%5E2%28x%29%7D)
subsituting
![csc^2(x)(1+cos^2(x))=](https://tex.z-dn.net/?f=csc%5E2%28x%29%281%2Bcos%5E2%28x%29%29%3D)
distributing
![2csc^2(x)-\frac{csc^2(x)}{csc^2(x)}=](https://tex.z-dn.net/?f=2csc%5E2%28x%29-%5Cfrac%7Bcsc%5E2%28x%29%7D%7Bcsc%5E2%28x%29%7D%3D)
is the simplified expression