<s>[INST] 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?
 [/INST] $\sin^2(x)-\cos^2(x)$ does not equal $-1$ or $1$.
$$\sin^2(x)+\cos^2(x)=1,$$
So $\sin^2(x)-\cos^2(x)$ is equal to:
$$\sin^2(x)-(1-\sin^2(x))= 2\sin^2(x)-1,$$ or alternatively, 
$$1-\cos^2(x)-\cos^2(x)= 1-2\cos^2(x).$$
</s>