If $x \in (0,\frac \pi6)$, then using calculus prove that $x\csc x<\frac \pi3$
My attempt:
let $f(x)=\csc x$$$\implies f'(x)=-\frac{\cos{x}}{\sin^2 x}$$ which is less than $0$ for all $x\in (0,\pi/6)$
so, $f'(x)<0$$\implies f(x)$ is decreasing function.
(as we know decreasing function reverses inequality)
$$\csc(0)>\csc(x)>2$$ (notice that inequality is reversed)and since $x$ is positive $$2x<x\csc x<\infty$$but this violates the question and i am definitely wrong somewhere but i don't know where!
Any Help will be appreciated :-D