Prove that $\frac{\tan{x}}{\tan{y}}>\frac{x}{y} : \forall (0<y<x<\frac{\pi}{2})$ Prove that $\frac{\tan{x}}{\tan{y}}>\frac{x}{y} : \forall (0<y<x<\frac{\pi}{2})$. My try, considering $f(t)=\frac{\tan{x}}{\tan{y}}-\frac{x}{y}$ and derivating it to see whether the function is increasing in the given interval. I should be sure that $\lim_{x,y\rightarrow0}\frac{\tan{x}}{\tan{y}}-\frac{x}{y}\geq0$ for the previous derivative check to be useful, which I'm not yet, but I'm assuming it's $0$ since I'd say that since both $x,y$ approach $0$ equally then the quotient of both their tangents and themselves is $1$, hence the substraction being $0$. However, the trouble arrives at the time of derivating it because of the 2 variables, I'm not sure if I have to fix one and derivate in terms of the other one, or what to do. I have to say I'm currently coursing a module on real single-variable analysis, so it can't have anything to do with multivariable analysis.

Consider the function

$$f(x) = \frac{\tan x}{x}.$$

You need to show that function is increasing on $(0,\pi/2)$.

If you differentiate that, you find

$$f'(x) = \frac{x(1+\tan^2 x) - \tan x}{x^2},$$

and you need to show $x(1+\tan^2 x) > \tan x$ for $x\in (0,\pi/2)$. A little trick helps showing that.