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!
 A: Hint:
separation of $(a,b)$.
details:


*

*for the first one, consider
$$g(x) = \frac x{\sin x}
\\
g'(x) = \frac 1{\sin x} - \frac {x\cos x}{\sin^2 x}=
\frac{\cos x}{\sin^2 x} (\tan x - x)> 0
$$
so $g$ is strictly increasing, and $$
\frac b{\sin b}< \frac a{\sin a}\iff \frac{\sin a}{\sin b} <\frac ab
$$

*For the second one, do the same thing with
$$
h(x) = \frac{\tan x}x
$$

A: 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 $\frac{\tan a}a>\frac{\tan b}b\Longrightarrow\frac ab<\frac{\tan a}{\tan b}$.
