Prove that $\frac{1}{\sin t} - \frac{1}{t}$ is increasing on $(0,\pi/2)$. I would like to obtain a rigorous proof of the fact that
$$
\frac{1}{\sin t} - \frac{1}{t}
$$
is increasing on $(0,\pi/2)$. I attempted the usual by taking the derivative and seeing if it's positive, however, this led me nowhere enlightening. Is it possibly related to the fact that $\sin x \leq x$ for all $0 \leq x \leq \pi/2$? This isn't a homework problem, so I would not mind a full solution. Thanks in advance!
 A: Taking the derivative, you want to show that
$$ \frac{ - \cos t } { \sin^2 t } + \frac{1}{t^2 } \geq 0  \Leftrightarrow \tan t \sin t \geq t^2 $$
There are many ways to prove this, like in this other question.
A: $$
\text{derivative} = \frac{-\cos t}{\sin^2 t} + \frac{1}{t^2} =\frac{-t^2\cos t + \sin^2 t}{\text{something positive}}
$$
$$
\text{the numerator} = -t^2\left(1 - \frac{t^2}{2}+\text{higher-degree terms}\right) + \left(t^2 - \frac{t^4}3 + \text{higher-degree terms}\right)
$$
$$
= \frac{t^4}{12} + \text{higher-degree terms}
$$
This shows at least that the derivative is positive if $t$ is close enough to $0$.  To get all the way up to $\pi/2$ will take more work.  Showing that the second derivative is positive will do it.
A: All inequalities are meant to be taken over $(0,\pi/2)$. The inequalities involving Taylor polynomials can be proven using the remainder which is either negative or positive in $(0,\pi/2)$.
Note that we can try and prove $$\left(\frac{\sin t}t\right)^2>\cos t$$
We use $1-\dfrac {t^2}2+\dfrac {t^4}{24}>\cos t$, so we aim to prove $${\left( {\frac{{\sin t}}{t}} \right)^2} > 1 - \frac{{{t^2}}}{2} + \frac{{{t^4}}}{{24}}$$
This is equivalent to showing $${\sin ^2}t > {t^2} - \frac{{{t^4}}}{2} + \frac{{{t^6}}}{{24}}$$ But $$\sin^2 t> t^2- \frac{t^4} 3$$
Then we have to show $${t^2} - \frac{{{t^4}}}{3} > {t^2} - \frac{{{t^4}}}{2} + \frac{{{t^6}}}{{24}}$$ or what is the same $$\frac{{{t^4}}}{6} > \frac{{{t^6}}}{24}$$
to be positive on $(0,\pi/2)$. And this is true.
A: You can use Maple to examine the claim's correctness as well. 
[> f:=t->1/sin(t)-1/t:
[> solve(diff(f(t),t)>0,t);

       RealRange(Open(-Pi), Open(0)), RealRange(Open(0), Open(Pi))

