What's $tg$ ? @below Oh I see. It's in radians right?

Dem0nfang wrote: What's $tg$ ? $\tan 1$

Pinko wrote: Prove the following inequality: $tan \, 1>\frac{3}{2}$. Using Taylor expansion of $\tan x$ might help

Pinko wrote: Prove the following inequality: $tan \, 1>\frac{3}{2}$. Over $[0,\frac{\pi}2)$, $\tan x$ is convex and so above any of its tangent lines. Tangent line at $x=\frac{\pi}3$ is $Y=4(X-\frac{\pi}3)+\sqrt 3$ So $\tan 1>4(1-\frac{\pi}3)+\sqrt 3>4(1-1.05)+1.7=1.5$ (using $\pi<3.15$ and $\sqrt 3>1.7$)