Originally Posted by
CaptainBlack Work in radians and expand as a Taylor series about $\displaystyle 60^{\circ}=\pi/3 {\rm{rad}}$
$\displaystyle
f(x+\varepsilon)=f(x)+\varepsilon~\left. \frac{df}{dx}\right|_{x}+\frac{\varepsilon^2}{2}~\ left. \frac{d^2f}{dx^2}\right|_{x}+..
$
Now:
$\displaystyle \frac{d}{dx}\tan(x)=(\sec(x))^2=1+(\tan(x))^2$,
and
$\displaystyle \frac{d^2}{dx^2}\tan(x)=2 (\sec(x))^2 \tan(x)=2\left[\frac{d}{dx}\tan(x)\right] \tan(x)$.
Now all you need is that $\displaystyle \tan(\pi/3)=\sqrt{3}$, and the $\displaystyle \varepsilon$ corresponding to $\displaystyle 1^{\circ}$ is $\displaystyle \varepsilon=\pi/180$
RonL