<s>[INST] 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 $\sec x -\sin x$
 [/INST] $$\scriptsize\frac {\tan^2x+\color{blue}{\cos^2x}}{\sin x+\sec x}=\frac {\overbrace{\tan^2x+\color{blue}1}^{\sec^2 x}\color{blue}{-\sin^2x}}{\sin x+\sec x}=\frac {\sec^2x-\sin^2x}{\sec x+\sin x}=\frac{(\sec x -\sin x)(\sec x+\sin x)}{\sec x+\sin x}=\sec x-\sin x$$
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