<s>[INST] Prove $\frac{\sin(a)}{\sin(b)}<\frac{a}{b}<\frac{\tan(a)}{\tan(b)}$ for $0not sure how to approach the following $\frac{\sin(a)}{\sin(b)}<\frac{a}{b}<\frac{\tan(a)}{\tan(b)}$ for $0<b<a<\pi/2$. Hints would be appreciated!
 [/INST] Note that the function $\frac{\sin x}x$ is decreasing:
$$\left(\frac{\sin x}x\right)'=\frac{x\cos x-\sin x}{x^2}=\frac{\cos x}{x^2}(x-\tan x)<0,$$
because $\cos x>0$ and $\tan x>x$ on $(0,\frac\pi2)$.
So for $\frac\pi2>a>b>0$ we have $\frac{\sin a}a<\frac{\sin b}b\Longrightarrow\frac{\sin a}{\sin b}<\frac ab$.
Additionally $\frac{\tan x}x$ is increasing:
$$\left(\frac{\tan x}x\right)'=\frac{\frac{x}{\cos^2 x}-\tan x}{x^2}=\frac{x-\sin x\cos x}{x^2\cos^2 x}=\frac{2x-\sin(2x)}{2x^2\cos^2x}>0,$$
because $\sin x<x$ on $(0,\infty)$.
So for $\frac\pi2>a>b>0$ we have $\frac{\tan a}a>\frac{\tan b}b\Longrightarrow\frac ab<\frac{\tan a}{\tan b}$.
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