How to show that when $t\in\left[0,\frac{π}{2}\right]$, the derivative of $\frac{1}{\sin t}-\frac{1}{t}$ is an increasing function? Thanks!
It is easy to prove that $f(t)=\frac{1}{\sin t}-\frac{1}{t}$ is an increasing function on $[0,\frac{\pi}{2}]$. But it seems difficult to prove that the first derivative of $f(t)$ is also an increasing function on $[0,\frac{\pi}{2}]$.
A possible approach is to notice that all the coefficients of the Taylor series of $f(t)=\frac{1}{\sin t}-\frac{1}{t}$ at the origin are non-negative, hence $f(t)$ is a convex function. The term $-\frac{1}{t}$ cancels the pole of $\frac{1}{\sin t}$ at the origin, the radius of convergence of the previous Taylor series is $\frac{\pi}{2}$ (by simply locating the closest singularity to the origin) and $f(t)$ is an odd function. Since $$ \text{Res}\left(\frac{1}{\sin t},t=k\pi\right) = (-1)^k\tag{1}$$ we have: $$ \frac{1}{\sin t}-\frac{1}{t} = \sum_{k\geq 1}(-1)^k\left(\frac{1}{t-k\pi}+\frac{1}{t+k\pi}\right) = \sum_{k\geq 1}\frac{(-1)^k 2t}{t^2-k^2\pi^2} \tag{2}$$ hence: $$ \frac{t}{\sin t}-1 = 2\sum_{k\geq 1}\frac{(-1)^{k+1}\frac{t^2}{\pi^2 k^2}}{1-\frac{t^2}{k^2\pi^2}}= 2\sum_{m\geq 1}t^{2m}\sum_{k\geq 1}\frac{(-1)^{k+1}}{k^{2m}\pi^{2m}}=2\sum_{m\geq 1}\frac{(4^m-2)\zeta(2m)}{(2\pi)^{2m}} t^{2m}\tag{3}$$ and the claim follows from $\frac{(4^m-2)\zeta(2m)}{(2\pi)^{2m}}\geq 0$.