Ask Me Help Desk - (cos x/sec x)-(cos x/tan^2 x)= cot^2 x?

I looked upon it, but I think that it's something wrong with the exercise.

$\frac{\cos x}{\sec x} - \frac{\cos x}{\tan ^2 x}=\cot ^2 x$

$\sec x= \frac{1}{\cos x}$

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

$\cos ^2 x - \frac{\cos ^3 x}{\sin ^2 x}=$

$\frac{\cos ^2 x \sin ^2 x - \cos ^3 x}{\sin ^2 x}=$

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

$\frac{\cos ^2 x}{\sin ^2 x}= \cot ^2 x$

$\cot ^2 x (\sin ^2 - \cos x)$

if there had been $\sin ^ 2 x + \cos ^2 x$ instead of $\sin ^2 x - \cos x$ it would have been true, but maybe I did some mistakes. On the whole, the idea is to use the formulas and calculate.