Thanks,
Ackbach for reminding me that my post isn't complete and isn't very convincing.
I think I should have mentioned first that $a$ is defined for all real numbers.
Next, I wanted a re-post to fix things right...
If we let $x=a-135^{\circ}$, where $a\in R$, we see that we have:
[TABLE="class: grid, width: 800"]
[TR]
[TD]$\begin{align*}\sin x&=\sin (a-135^{\circ})\\&=\dfrac{-\sin a-\cos a}{\sqrt{2}}\end{align*}$[/TD]
[TD]$\begin{align*}\cos x&=\cos (a-135^{\circ})\\&=\dfrac{-\cos a+\sin a}{\sqrt{2}}\end{align*}$[/TD]
[TD]$\begin{align*}\sin x+\cos x&=\dfrac{-\sin a-\cos a}{\sqrt{2}}+\dfrac{-\cos a+\sin a}{\sqrt{2}}\\&=-\sqrt{2}\cos a\end{align*}$[/TD]
[/TR]
[TR]
[TD]$\begin{align*}\tan x&=\tan(a-135^{\circ})\\&=\dfrac{\sin a+\cos a}{\cos a-\sin a}\end{align*}$[/TD]
[TD]$\begin{align*}\cot x&=\cot(a-135^{\circ})\\&=\dfrac{\cos a-\sin a}{\sin a+\cos a}\end{align*}$[/TD]
[TD]$\begin{align*}\tan x+\cot x&=\dfrac{\sin a+\cos a}{\cos a-\sin a}+\dfrac{\cos a-\sin a}{\sin a+\cos a}\\&=\dfrac{2}{\cos^2 a-\sin^2 a}\end{align*}$[/TD]
[/TR]
[TR]
[TD]$\begin{align*}\sec x&=\dfrac{1}{\cos x}\\&=-\dfrac{\sqrt{2}}{\sin a+\cos a}\end{align*}$[/TD]
[TD]$\begin{align*}\csc x&=\dfrac{1}{\sin x}\\&=\dfrac{\sqrt{2}}{\sin a-\cos a}\end{align*}$[/TD]
[TD]$\begin{align*}\sec x+\csc x&=-\dfrac{\sqrt{2}}{\sin a+\cos a}+\dfrac{\sqrt{2}}{\sin a-\cos a}\\&=\dfrac{2\sqrt{2}\cos a}{\sin^2 a-\cos^2 a}\end{align*}$[/TD]
[/TR]
[/TABLE]
$\begin{align*}\therefore|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|&=|-\sqrt{2}\cos a+\dfrac{2}{\cos^2 a-\sin^2 a}+\dfrac{2\sqrt{2}\cos a}{\sin^2 a-\cos^2 a}|\\&=|-\sqrt{2}\cos a+\dfrac{2}{\cos^2 a-\sin^2 a}-\dfrac{2\sqrt{2}\cos a}{\cos^2 a-\sin^2 a}|\\&=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{\cos^2 a-\sin^2 a}|\\&=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{2\cos^2 a-1}|\\&=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{(\sqrt{2}\cos a-1)(\sqrt{2}\cos a+1)}|\\&=|-\sqrt{2}\cos a+\dfrac{2(1-\sqrt{2}\cos a)}{(\sqrt{2}\cos a-1)(\sqrt{2}\cos a+1)}|\\&=|-\sqrt{2}\cos a-\dfrac{2}{\sqrt{2}\cos a+1}|\\&=|-\left(\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \right)|\\&=|\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1}|\end{align*}$
[TABLE="class: grid, width: 800"]
[TR]
[TD]If we have $(\sqrt{2}\cos a+1)$ and $\dfrac{2}{\sqrt{2}\cos a+1}$ where $\sqrt{2}\cos a+1>0$, by applying AM-GM inequality to these two terms yields
$\dfrac{(\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1}}{2}\ge \sqrt{(\sqrt{2}\cos a+1)\left(\dfrac{2}{\sqrt{2}\cos a+1} \right)}\ge \sqrt{2}$, or simply
$(\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1}\ge 2 \sqrt{2}$
$\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \ge 2 \sqrt{2}-1$[/TD]
[TD]If we have $(\sqrt{2}\cos a+1)$ and $\dfrac{2}{\sqrt{2}\cos a+1}$ where $\sqrt{2}\cos a+1<0$ which also implies $\sqrt{2}\cos a<0$, by applying AM-GM inequality to these two terms yields
$\dfrac{-\left((\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1} \right)}{2}\ge \sqrt{(\sqrt{2}\cos a+1)\left(\dfrac{2}{\sqrt{2}\cos a+1} \right)}\ge \sqrt{2}$, or simply
$-\left((\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1} \right)\ge 2 \sqrt{2}$
$(\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1} \le -2 \sqrt{2}$
$\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \le -2 \sqrt{2}-1$[/TD]
[/TR]
[/TABLE]
Now if we set $\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} = 2 \sqrt{2}-1$, this gives us $\sqrt{2}\cos a=\dfrac{\sqrt{2}-1}{\sqrt{2}}<1$, which means what we set to be true has solution(s), then we know we're dealing with tight inequality here, and hence, we can conclude at this point the minimum value for $$|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$$ is $2 \sqrt{2}-1$.
Does this look okay now,
Ackbach?(Thinking)
