Prove that $\frac{\tan x}{x}>\frac{x}{\sin x}, x\in(0,\pi/2)$ 
Prove that $$\frac{\tan x}{x}>\frac{x}{\sin x},\;\;\; x\in(0,\pi/2).$$

My work
I formulated $$f(x)=\tan x \sin x - x^2$$ in hope that if $f'(x)>0$ i.e. monotonic then I can conclude for $x>0, f(x)>f(0)$ and hence, prove the statement.
However, I got $$f'(x)=\sin x + \sec x \tan x -2x, $$ where I am unable to conclude if $f'(x)>0.$
I also found $$f''(x)=\cos x + 2\sec^3x-\sec x-2,$$
$$f'''(x)=-\sin x (1-6\sec^4x+\sec^2x).$$
But I am not able conclude the sign of any of the higher derivatives either. Am I doing something wrong? Or is there some other way?

We need to prove that
$$\frac{\sin^2x}{\cos{x}}>x^2$$ or $f(x)>0$, where
$$f(x)=\frac{\sin{x}}{\sqrt{\cos{x}}}-x.$$
Now, let $\cos{x}=t^2$, where $0<t<1$.
Thus, $$f'(x)=\frac{1+\cos^2x}{2\sqrt{\cos^3x}}-1=\frac{(1-t)(1+t+t^2-t^3)}{2t^3}>0,$$ which says $f(x)>f(0)=0$ and  we are done!
