<s>[INST] Compare $\arcsin (1)$ and $\tan (1)$ 
Which one is greater: $\arcsin (1)$ or $\tan (1)$? 

How to find without using graph or calculator?
I tried using $\sin(\tan1)\leq1$, but how do I eliminate the possibility of an equality without using the calculator?
 [/INST] $\arcsin(1)=\frac{\pi}{2}$ while from the Weierstrass product for the cosine function we have
$$ \tan(1) = \sum_{n\geq 0}\frac{8}{(2n+1)^2 \pi^2-4 }=2\sum_{n\geq 0}\left[\frac{1}{(2n+1)\pi-2}-\frac{1}{(2n+1)\pi+2}\right]$$
such that an effective integral representation of $\tan(1)$ through an almost-Gaussian integral is
$$\tan(1)=\int_{\mathbb{R}}\frac{\sinh(2x)}{\sinh(\pi x)}\,dx.$$
As an alternative, by expanding $\frac{8}{(2n+1)^2\pi^2-4}$ as a geometric series we get
$$ \tan(1) = 2 \sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)}{\pi^{2m}} $$
which is equivalent to the previous integral representation via $\zeta(2m)=\frac{1}{(2m-1)!}\int_{0}^{+\infty}\frac{z^{2m-1}}{e^z-1}\,dz$.
In order to prove that $\tan(1)<\frac{\pi}{2}$ it is enough to show that $\tan\left(\frac{1}{2}\right)<\frac{\pi}{2+\sqrt{\pi^2+4}}$, since $\tan(2z)=\frac{2\tan z}{1-\tan^2 z}$. $\tan\left(\frac{1}{2}\right)$ has a fast-convergent series representation
$$ \tan\left(\tfrac12\right)=4\sum_{m\geq 1}\frac{\zeta(2m)}{\pi^{2m}}\left(1-\frac{1}{4^m}\right) $$
which allows to state
$$ \tan\left(\tfrac12\right) < 4\sum_{m= 1}^{3}\frac{\zeta(2m)}{\pi^{2m}}\left(1-\frac{1}{4^m}\right)+4\,\zeta(8)\sum_{m \geq 4}\frac{1}{\pi^{2m}}\left(1-\frac{1}{4^m}\right)$$
or
$$ \tan\left(\tfrac12\right)<\frac{131}{240}+\frac{\pi ^2 \left(85 \pi^2-21\right)}{50400 \left(\pi^2-1\right) \left(4 \pi^2-1\right)}$$
(this is extremely accurate). The proof is finished by exploiting $\frac{227}{23}<\pi^2<\frac{79}{8}$ which follows from the study of the Beuker-like integrals
$$ \iint_{(0,1)^2}\frac{x^m(1-x)^m y^n (1-y)^n}{1-xy}\,dx\,dy.$$

An alternative approach. The Shafer-Fink inequality gives that for any $x\in(0,3/2)$
$$ \tan(x) < \frac{3x+2x\sqrt{9-3x^2}}{9-4x^2} $$
holds, so by the duplication formula for the tangent function the sharper inequality
$$ \tan(x) < \frac{4 x\left(9-x^2\right) \left(3+\sqrt{36-3 x^2}\right)}{324-117 x^2+7 x^4-6 x^2 \sqrt{36-3 x^2}} $$
holds too. This gives
$$ \tan(1) < \frac{32 \left(3+\sqrt{33}\right)}{214-6 \sqrt{33}}=\frac{4}{697} \left(105+29 \sqrt{33}\right) $$
so it is enough to show that $\pi>\frac{32(3+\sqrt{33})}{107-3\sqrt{33}}$, which is a pretty loose inequality.
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