Simplification of Trigo expression 
Simplify $$\frac{\tan^2 x+\cos^2 x}{\sin x+ \sec x}$$

My attempt, 
$$=\frac{(\cos^2x+\frac{\sin^2x}{\cos^2x})}{\frac{\sin x \cos x+1}{\cos x}} $$
$$=\frac{\cos^4 x+\sin^2 x}{\cos^2 x}\cdot \frac{\cos x}{\sin x \cos x+1}$$
$$=\frac{\cos^4 x+\sin^2 x}{\cos x(\sin x \cos x+1)}$$
I'm stuck at here. The given answer is $\sec x -\sin x$

Hint
Use $\tan^2x=\sec^2x -1$
$$\frac{\tan^2 x+\cos^2 x}{\sin x+ \sec x}=\frac{(\sec^2 x-1)+\cos^2 x}{\sin x+ \sec x}=\frac{\sec^2 x-\sin^2 x}{\sin x+ \sec x}$$
