Prove that $\frac{\sin(a)}{\sin(b)} < \frac{a}{b} < \frac{\tan(a)}{\tan(b)}$ where $0 < b < a < \frac{\pi}{2}$

Question

Prove the following:

$\frac{\sin(a)}{\sin(b)} < \frac{a}{b} < \frac{\tan(a)}{\tan(b)}$ where $0 < b < a < \frac{\pi}{2}$

Hello everyone, I am trying to create some sort of function or maybe see an application of the mean value theorem, but I am just not getting it. Any help would be greatly appreciated!

Answer 1

Hint: Consider $\frac{\sin x}{x}$ and $\frac{\tan x}{x}$. You need to show that they are decreasing / increasing on $(0,\frac{\pi}{2})$.