Using Mean Value Theorem, Prove that ${\tan x\over x}>{x\over\sin x}$

We need to prove that $f(x)>0$, where $f(x)=\frac{\sin{x}}{\sqrt{\cos{x}}}-x$.

But $f'(x)=\frac{\cos x\sqrt{\cos x}+\frac{\sin^2x}{2\sqrt{\cos x}}}{\cos{x}}-1=\frac{(\sqrt{\cos^3{x}}-1)^2+\cos^2x(1-\cos{x})}{2\sqrt{\cos^3x}}>0$.

Thus, $f(x)>f(0)=0$.

Done!