Evaluating $(\frac{\cos x}{1-\sin x})^2$ $(\dfrac{\cos x}{1-\sin x})^2$  
$f\;'(x)= 2(\dfrac{\cos x}{1-\sin x}) \times (\dfrac{-\sin x+\sin^2x-\cos^2x}{(1-\sin x)^2})$  
Does $\sin^2x-\cos^2=1$? or $-1$?  Then it could factor with the bottom and the answer would be $\dfrac{2\cos x}{(1-\sin x)^2}$. Is this right?

Using $\sin^2 x + \cos^2 x=1$, 
$$f\;'(x)= 2(\dfrac{\cos x}{1-\sin x}) \times (\dfrac{-\sin x(1-\sin x)+\cos^2x}{(1-\sin x)^2})= 2(\dfrac{\cos x}{1-\sin x}) \times (\dfrac{1-\sin x}{(1-\sin x)^2})=\frac{2\cos x}{(1-\sin x)^2}.$$  
