First, let's simplify the given expression to contain just $\sin x$ and $\cos x$: \begin{align*} f(x)&=|\sin x + \cos x + \tan x + \cot x + \sec x + \csc x|\\ &=\Big|\sin x + \cos x + \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} +\frac{1}{\cos x} + \frac{1}{\sin x}\Big|\\ &=\Big|\frac{\sin^2 x\cos x + \cos^2 x\sin x + \sin^2x + \cos^2 x + \sin x + \cos x}{\sin x\cos x}\Big|\\ &=\Big|\frac{\sin x\cos x(\sin x + \cos x) + 1 + (\sin x +\cos x)}{\sin x\cos x}\Big|\\ &=\Big|(\sin x + \cos x) +\frac{ 1 + (\sin x +\cos x)}{\sin x\cos x}\Big| \end{align*} Let $y=\sin x + \cos x$. Then $$y^2 = 1+2\sin x\cos x \implies \sin x\cos x = \frac{1}{2}(y^2-1)$$ Therefore $f(x)$ is equivalent to \begin{align*} f(y)=\Big|y + \frac{2(1+y)}{y^2 - 1}\Big|=\Big|y+\frac{2}{y-1}\Big|=\Big|(y-1)+\frac{2}{y-1}+1\Big| \end{align*} The range of $y$ is $[-\sqrt{2},\sqrt{2}]$ because $$y=\sin x + \cos x = \sqrt{2}\sin(x + 45^\circ)$$ i) If $(y-1) > 0$, or equivalently, $y\in (1, \sqrt{2}]$, then $$(y-1)+\frac{2}{y-1}+1\ge 2\sqrt{(y-1)\cdot\frac{2}{y-1}}+1=2\sqrt{2}+1 \ge 0$$ $$\implies f(y) \ge 2\sqrt{2}+1$$ ii) If $(y-1) < 0$, or equivalently, $y\in [-\sqrt{2}, 1)$, then $$(y-1)+\frac{2}{y-1}+1 \ge -2\sqrt{(1-y)\cdot\frac{2}{1-y}}+1 = -2\sqrt{2}+1 < 0$$ $$f(y)\ge 2\sqrt{2}-1$$ Therefore, in conclusion, we find the minimal value of given expression equals $\boxed{2\sqrt{2}-1}$ which is reachable when $$1-y=\frac{2}{1-y}, y < 1 \implies y=1-\sqrt{2}\implies \sin(x+45^\circ)=\frac{\sqrt{2}}{2}-1$$