I feel terrible for making yet another thread, but I've been stuck on this for 30 mins:
LS=cos(2x)/(1-sin2x)
RS=(cosx+sinx)/(cosx-sinx)
Hello, youngb11!
Another approach . . .
$\displaystyle \dfrac{\cos 2x}{1-\sin2x} \;=\;\dfrac{\cos x+\sin x}{\cos x-\sin x}$
Multiply the right side by $\displaystyle \frac{\cos x - \sin x}{\cos x - \sin x}$
. . $\displaystyle \displaystyle \frac{\cos x + \sin x}{\cos x - \sin x}\cdot\frac{\cos x - \sin x}{\cos x - \sin x} \;=\;\frac{\cos^2\!x - \sin^2\!x}{\cos^2\!x - 2\cos x\sin x + \sin^2\!x} $
. . . . $\displaystyle \displaystyle =\;\frac{\overbrace{\cos^2\!x - \sin^2\!x}^{\text{This is }\cos2x}}{\underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}} - \underbrace{2\sin x\cos x}_{\text{This is }\sin2x}} \;=\;\frac{\cos2x}{1-\sin2x} $
Just use the identities I gave you otherwise.
$\displaystyle \dfrac{\cos 2x}{1-\sin2x} = \dfrac{\cos^2(x) - \sin^2(x)}{1 - 2\cos(x)\sin(x)}$
Here, the trick is to spot that the numerator is a difference of two squares and that the numerator can be a square by using the Pythagorean identity, $\displaystyle \cos^2(x) + \sin^2(x) = 1$
This said, you get:
$\displaystyle \dfrac{\cos^2(x) - \sin^2(x)}{1 - 2\cos(x)\sin(x)} = \dfrac{(\cos(x) - \sin(x))(\cos(x) + sin(x))}{\cos^2(x) + \sin^2(x) - 2\cos(x)\sin(x)}$
You then factor the denominator;
$\displaystyle \dfrac{(\cos(x) - \sin(x))(\cos(x) + sin(x))}{\cos^2(x) + \sin^2(x) - 2\cos(x)\sin(x)} = \dfrac{(\cos(x) - \sin(x))(\cos(x) + sin(x))}{(\cos(x) - \sin(x))^2}$
Some things cancel out, leaving the RS.
You can also see that it's the reverse process that Soroban posted