1. ## Proving Trig. Identity

The original problem is $\displaystyle (1-cos(x))/(1+cos(x))=(csc(x)-cot(x))^2$

I start by solving from the right side. First I expand the binomial and then I take reciprocal of both csc and cot. At that point I get $\displaystyle ((cos(x) -1)^2)/(1-cos(x))$ After that, I have no idea how to get the left side honestly. What do I do from here? Thanks

2. $\displaystyle (\csc{x} - \cot{x})^2 = \left(\frac{1}{\sin{x}} - \frac{\cos{x}}{\sin{x}}\right)^2$

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

$\displaystyle = \frac{(1 - \cos{x})^2}{\sin^2{x}}$

$\displaystyle = \frac{(1 - \cos{x})^2}{1 - \cos^2{x}}$

$\displaystyle = \frac{(1 - \cos{x})^2}{(1 - \cos{x})(1 + \cos{x})}$

$\displaystyle = \frac{1 - \cos{x}}{1 + \cos{x}}$.

3. Originally Posted by Altermeris
The original problem is $\displaystyle (1-cos(x))/(1+cos(x))=(csc(x)-cot(x))^2$

I start by solving from the right side. First I expand the binomial and then I take reciprocal of both csc and cot. At that point I get $\displaystyle ((cos(x) -1)^2)/(1-cos(x))$ After that, I have no idea how to get the left side honestly. What do I do from here? Thanks
from LHS
show that: $\displaystyle \frac{1-cos(x)}{1+cos(x)}=(csc(x)-cot(x))^2$

$\displaystyle \frac{1-cos(x)}{1+cos(x)}=\frac{1-cos(x)}{1+cos(x)} X \frac{1-cos(x)}{1-cos(x)}$

$\displaystyle \frac{1-cos(x)}{1+cos(x)}=\frac{(1-cos(x))^2}{1-cos^2(x)}$

$\displaystyle \frac{1-cos(x)}{1+cos(x)}=\frac{(1-cos(x))^2}{sin^2(x)}$

$\displaystyle \frac{1-cos(x)}{1+cos(x)}=(\frac{1-cos(x)}{sin(x)})^2$

$\displaystyle \frac{1-cos(x)}{1+cos(x)}=(\frac{1}{sin(x)}-\frac{cos(x)}{sin(x)})^2$

$\displaystyle \frac{1-cos(x)}{1+cos(x)}=(csc(x)-cot(x))^2$

-- hope it'll help --