Simplify Trigonometric Expression

I need to simplify this expression

$$\frac{\tan\theta + \cos\theta}{\cos\theta \sin\theta}-\frac{1}{\cos^2\theta}$$

I got $\frac{1}{\cos \theta \sin\theta}$ as an answer but im not sure if this is correct. Can anyone give the steps on how to solve this?

Hint: $$\frac{\tan\theta + \cos\theta}{\cos\theta \sin\theta}-\frac{1}{\cos^2\theta}=\frac{\cos \theta(\tan\theta+\cos\theta)-\sin \theta}{\cos^2\theta \sin\theta}$$
and $\cos\theta \tan \theta=\sin\theta$. The answer is $\frac{1}{\sin\theta}$.
\eqalign{ & \,\,\,\,{{\tan \theta + \cos \theta } \over {\cos \theta \sin \theta }} - {1 \over {{{\cos }^2}\theta }} = {{{{\sin \theta } \over {\cos \theta }} + \cos \theta } \over {\cos \theta \sin \theta }} - {1 \over {{{\cos }^2}\theta }} \cr & = {{\sin \theta + {{\cos }^2}\theta } \over {{{\cos }^2}\theta \sin \theta }} - {1 \over {{{\cos }^2}\theta }} \cr & = {{\sin \theta + {{\cos }^2}\theta - \sin \theta } \over {{{\cos }^2}\theta \sin \theta }} \cr & = {1 \over {\sin \theta }} \cr}
$$\dfrac{\tan\theta}{\cos\theta\sin\theta}=\dfrac{\sin\theta}{\cos^2\theta\sin\theta}=\dfrac1{\cos^2\theta}$$