<s>[INST] Small angle approximations - different answers I would like to approximate $$\frac{\cos^2{x}}{\sin(x) \tan(x)}$$ using the small angle approximations. 
Throughout I will use $\sin(x) \approx x$, $\tan(x) \approx x$, $\cos(x) \approx 1 - \frac{x^2}{2}$.
Method 1: 
$$\frac{\cos^2{x}}{\sin(x) \tan(x)} \approx \frac{\left(1-\frac{x^2}{2}\right)^2}{x^2} = \frac{1}{4}(x^2 - x^{-2} - 1)$$
Method 2: 
$$\frac{\cos^2{x}}{\sin(x) \tan(x)} = \frac{\cos^3{x}}{\sin^2(x)} \approx \frac{1}{x^2} \left(1 - \frac{x^2}{2}\right)^3$$
which when expanded out gives for example $-\frac{3}{2} \neq - \frac{1}{4}$ as the constant term. 
Why the differences? Which is the better way to approximate this and why?
 [/INST] Because you are using less terms in your first expansion. What happens if you take $\tan(x) \approx \frac{x}{1 -x^2/2}$? It gives the second expansion. So the first one is a weaker approximation. In the first your are not keeping all terms of order $x^2$, but you do in the second. Note that the leading order terms for small $x$ are $x^{-2}$ which are the same. 
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