<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] I try to calculate $\tan 1 <1.57<\frac{\pi}{2}$
$\tan x =x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\frac{62x^9}{2835}+o(x^9)$
$|\tan x-x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\frac{62x^9}{2835}|<\frac{62x^9}{2835},x\in[0,1]$
So:$\tan 1<1+\frac{1}{3}+\frac{2}{15}+\frac{17}{315}+\frac{62}{2835}+\frac{62}{2835}<1.565<1.57<\frac{\pi}{2}=\arcsin 1$
Edits:
It seems that there just need four terms by under  @Ian reminding
$\tan x =x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\frac{62\xi^9}{2835},\xi\in[0,1]$
$\tan 1<1+\frac{1}{3}+\frac{2}{15}+\frac{17}{315}+\frac{62}{2835}<1.56<1.57<\frac{\pi}{2}=\arcsin 1$
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