To answer your question, yes, they are equivalent, since
[tex]\dfrac{\cos^2\theta}{\cos\theta-\cos\theta\sin\theta}=\dfrac{1-\sin^2\theta}{\cos\theta(1-\sin\theta)}=\dfrac{(1-\sin\theta)(1+\sin\theta)}{\cos\theta(1-\sin\theta)}=\dfrac{1+\sin\theta}{\cos\theta}[/tex]
Similar manipulations and simplifications will show that A, B, and D are all equivalent to the starting expression, so the answer would be C.
