Minimize the trigonometric expression.

[anemone]

Find the minimum value of \(\displaystyle |\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|\) for all real numbers \(\displaystyle x\).

 

[Prove It]

Find the minimum value of \(\displaystyle |\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|\) for all real numbers \(\displaystyle x\).

Well for starters, we know it can never be any less than 0. Whether or not 0 is the minimum is another story though...

 

[chisigma]

Find the minimum value of \(\displaystyle |\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|\) for all real numbers \(\displaystyle x\).

It is easy to verify that the sum of the derivatives of the functions $\sin x$ and $\cos x$, $\tan x$ and $\cot x$ , $\sec x$ and $\csc x$ vanishes for $x= \frac{\pi}{4}$...

Kind regards

$\chi$ $\sigma$

 

[anemone]

It is easy to verify that the sum of the derivatives of the functions $\sin x$ and $\cos x$, $\tan x$ and $\cot x$ , $\sec x$ and $\csc x$ vanishes for $x= \frac {\pi}{4}$...

Kind regards

$\chi$ $\sigma$

Thanks for participating in this problem but chisigma, I believe what you cited here, i.e. \(\displaystyle x=\frac{\pi}{4}\) only gives us the local minimum value but the question is asking for the absolute minimum value...

 

[Klaas van Aarsen]

Thanks for participating in this problem but chisigma, I believe what you cited here, i.e. \(\displaystyle x={\pi}{4}\) only gives us the local minimum value but the question is asking for the absolute minimum value...

To be fair, $\chi$ $\sigma$ ended his remark with dots, meaning that the real solution will come in a later post, which I'm sure he'll make.
If only the domain were limited to ($0, \frac \pi 2$)...
It's an interesting observation though.

Another interesting property is that the function is smooth in $x=\pi$ if we define 2 for its value.
However, this is neither a minimum nor a maximum.

Checking special values shows that $x=\frac 5 6 \pi$ gives us almost the answer.
But Wolfram says that x must be just a wee bit more...

Wolfram comes up with \(\displaystyle x = 2\arctan(1+\frac 1 {\sqrt 2} \mp \sqrt{\frac 1 2(5+4\sqrt2)}) \approx -1.083,\ 2.653 \pmod{2\pi}\) for a value of $1.82843$.

 

[chisigma]

It is easy to verify that the sum of the derivatives of the functions $\sin x$ and $\cos x$, $\tan x$ and $\cot x$ , $\sec x$ and $\csc x$ vanishes for $x= \frac{\pi}{4}$...

If we ignore the 'absolute value' the trigonometric function can be written as...


$\displaystyle f(x) = \frac{1 + \sin x + \cos x}{\sin x\ \cos x} + \sin x + \cos x$ (1)


... and observing (1) we discover that in $ 0 \le x \le 2 \pi$ f(*) has only two singularities in $x=0$ and $x=\frac{\pi}{2}$ because...

$\displaystyle \lim_{x \rightarrow \pi} \frac{1 + \sin x + \cos x}{\sin x\ \cos x} + \sin x + \cos x = -2$ (2)

... and...

$\displaystyle \lim_{x \rightarrow \frac{3}{2}\ \pi} \frac{1 + \sin x + \cos x}{\sin x\ \cos x} + \sin x + \cos x = -2$ (3)


The behavior of f(*) is represented in the figure...

... and it is evident that the minimum of the absolute value of f(*) is in the range $\displaystyle \frac{\pi}{2} \le x \le 2\ \pi$. The minima can be obtained forcing to zero the derivative of (1) but it seems a complicated procedure, so that the task is delayed to next posts...


Kind regards


$\chi$ $\sigma$

 

[anemone]

To be fair, $\chi$ $\sigma$ ended his remark with dots, meaning that the real solution will come in a later post, which I'm sure he'll make.
If only the domain were limited to ($0, \frac \pi 2$)...
It's an interesting observation though.

Hi I like Serena,

I think I should clear this up because when I made my previous remark, I did not intend to offend chisigma by saying his solution wasn't correct, but it was my mistake because I should have noticed chisigma is going to post for more to this thread that will eventually give us the final answer to this problem, I'm sorry if it came across that way.

Best regards,

anemone

 

[Klaas van Aarsen]

... and it is evident that the minimum of the absolute value of f(*) is in the range $\displaystyle \frac{\pi}{2} \le x \le 2\ \pi$. The minima can be obtained forcing to zero the derivative of (1) but it seems a complicated procedure, so that the task is delayed to next posts...

Setting the derivative of (1) to zero does not become too bad.
It boils down to $x=\frac \pi 4 \vee x=\frac 5 4 \pi \vee -\sin^2x\cos^2x + \sin x\cos x + \sin x + \cos x + 1 = 0$.
Where to go from there will come in a later post...

 

[Klaas van Aarsen]

I think I should clear this up because when I made my previous remark, I did not intend to offend chisigma by saying his solution wasn't correct, but it was my mistake because I should have noticed chisigma is going to post for more to this thread that will eventually give us the final answer to this problem, I'm sorry if it came across that way.

No harm done. Part of math is pointing out potential flaws in a reasoning.
Now that I think about it, I pointed out a couple of perceived flaws in a previous proof of yours.
I do hope you do not hold that against me.

 

[Ackbach]

This is a Putnam problem that I accidentally assigned to my AP Calculus AB students. Only one of them solved it. I'm guessing this solution is out on the Internet somewhere, because it involved two brilliant insights that I don't think my student actually dreamed up himself. Anyhoo, here's the solution:

Spoiler

Let $\xi= \sin(x)+ \cos(x)$. Then
$$g( \xi)= \frac{ \xi^{3}+ \xi+2}{ \xi^{2}-1}=f(x).$$
Minimizing $ \xi$ on its period of $[0,2 \pi)$ yields that it varies from $- \sqrt{2}$ to $ \sqrt{2}$. Hence, we have changed the problem to minimizing $g( \xi)$ on $[- \sqrt{2}, \sqrt{2}]$. The standard derivative approach yields $g'( \xi)=0$ when $\xi = 1 \pm \sqrt{2}$. But $1+ \sqrt{2} \not \in [- \sqrt{2}, \sqrt{2}]$, so we do not consider it. Hence, we are interested in when $ \xi=1- \sqrt{2}$. Note that $\sin(x)+ \cos(x)=1- \sqrt{2}$ when
$$ \sqrt{2} \left[ \sin \left( x+ \frac{ \pi}{4}\right)\right]=1- \sqrt{2},$$
so we need
$$ \sin \left( x+ \frac{ \pi}{4}\right)= \frac{ \sqrt{2}-2}{2},$$
or $g(1- \sqrt{2})=|1-2 \sqrt{2}|=2 \sqrt{2}-1.$
This is the minimum value of $f$ on the real line.

 

[MarkFL]

This is a Putnam problem that I accidentally assigned to my AP Calculus AB students. Only one of them solved it. I'm guessing this solution is out on the Internet somewhere, because it involved two brilliant insights that I don't think my student actually dreamed up himself. Anyhoo, here's the solution:

The OP is offline now, but she did share with me a link to the solution you gave that she found online a few hours ago. It is the same, almost word for word:

Spoiler

http://forumgeom.fau.edu/POLYA/ProblemCenter/TSPOLYA035A3.pdf

She does have her own solution though.

 

[anemone]

Thanks Mark for sharing the link on behalf of me!

My solution:

My approach is to find the minimum and maximum points of the function \(\displaystyle y=\sin x+\cos x+\tan x+\sec x+\csc x+\cot x\), and deduce the minimum value of the function \(\displaystyle y=|\sin x+\cos x+\tan x+\sec x+\csc x+\cot x|\) from them.

First thing that I noticed is the function \(\displaystyle y=\sin x+\cos x+\tan x+\sec x+\csc x+\cot x\) has a period of \(\displaystyle 2\pi\), thus, to solve for this problem, it suffices to just find the minimum and/or maximum points that with x values within \(\displaystyle 0\) and \(\displaystyle 2\pi\), inclusive.

Using differentiation to find the minimum and/or maximum point, I proceed as follows:

\(\displaystyle y=\sin x+\cos x+\tan x+\sec x+\csc x+\cot x\)

\(\displaystyle y'=\cos x-\sin x+\sec^2 x-\csc^2 x+\sec x \tan x+\csc x \cot x\)

\(\displaystyle y'=\cos x-\sin x+(1+\tan^2 x)-(1+\cot^2 x)+\sec x \tan x+\csc x \cot x\)

\(\displaystyle y'=\cos x-\sin x +\tan^2 x-\cot^2 x+\sec x \tan x+\csc x \cot x\)

\(\displaystyle y'=\cos x-\sin x +\tan x(\tan x+ \sec x)-\cot x(\cot x +\csc x)\)

\(\displaystyle y'=\cos x-\sin x +\frac{\sin x}{\cos x}\left(\frac{\sin x}{\cos x}+ \frac{1}{\cos x}\right)-\frac{\cos x}{\sin x}\left(\frac{\cos x}{\sin x} +\frac{1}{\sin x}\right)\)

\(\displaystyle y'=\cos x-\sin x +\frac{\sin x}{1-\sin x}-\frac{\cos x}{1-\cos x}\)

\(\displaystyle y'=\cos x-\sin x +\frac{\sin x-\cos x}{(1-\sin x)(1-\cos x)}\)

\(\displaystyle y'= \frac{(\sin x-\cos x)(\sin x \cos x-\cos x-\sin x)}{(1-\sin x)(1-\cos x)}\)

Set the derivative equal to zero and solve for x to find critical x values, we obtain:

\(\displaystyle (\sin x-\cos x)(\sin x \cos x-\cos x-\sin x)=0\)

\(\displaystyle \sin x-\cos x=0\) or \(\displaystyle \sin x \cos x-\cos x-\sin x=0\)

\(\displaystyle x=\frac{\pi}{4},\;\frac{5\pi}{4}\) or \(\displaystyle x=0.844611\pi,\;1.655389\pi\)

To determine whether the critical point is a minimum or maximum or neither, we choose a number just above and just below those critical values and get:

x	y'	This gives us a minimum point at (π/4, 6.242640687).
2π/9 (40°)	negative
π/4 (45°)	0
5π/18 (50°)	positive

x	y'	This gives us a minimum point at (5π/4, 6.242640687)
11π/9 (220°)	negative
5π/4 (225°)	0
23π/18 (230°)	positive

x	y'	This gives us a maximum point at (0.844611π, -1.828427125)
5π/6 (150°)	positive
0.844611π (152.03°)	0
31π/36 (155°)	negative

x	y'	This gives us a maximum point at (1.655389π, -1.828427125)
29π/18 (290°)	positive
1.655389π (297.97°)	0
5π/3 (300°)	negative

Therefore, I can conclude that the minimum value of \(\displaystyle y=|\sin x+\cos x+\tan x+\sec x+\csc x+\cot x|\) for all real x is 1.828427125.

 

[anemone]

Find the minimum value of \(\displaystyle |\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|\) for all real numbers \(\displaystyle x\).

I just saw another really brilliant way to tackle this problem and since I have a habit to share all that I know with the members of the forum, (ahem, I am quite generous myself), I will post the solution here without further ado.

If we let $x=a-135^{\circ}$, we see that we have:

$\begin{align*}\sin x&=\sin (a-135^{\circ})\\&=\dfrac{-\sin a-\cos a}{\sqrt{2}}\end{align*}$	$\begin{align*}\cos x&=\cos (a-135^{\circ})\\&=\dfrac{-\cos a+\sin a}{\sqrt{2}}\end{align*}$	$\begin{align*}\sin x+\cos x&=\dfrac{-\sin a-\cos a}{\sqrt{2}}+\dfrac{-\cos a+\sin a}{\sqrt{2}}\\&=-\sqrt{2}\cos a\end{align*}$
$\begin{align*}\tan x&=\tan(a-135^{\circ})\\&=\dfrac{\sin a+\cos a}{\cos a-\sin a}\end{align*}$	$\begin{align*}\cot x&=\cot(a-135^{\circ})\\&=\dfrac{\cos a-\sin a}{\sin a+\cos a}\end{align*}$	$\begin{align*}\tan x+\cot x&=\dfrac{\sin a+\cos a}{\cos a-\sin a}+\dfrac{\cos a-\sin a}{\sin a+\cos a}\\&=\dfrac{2}{\cos^2 a-\sin^2 a}\end{align*}$
$\begin{align*}\sec x&=\dfrac{1}{\cos x}\\&=-\dfrac{\sqrt{2}}{\sin a+\cos a}\end{align*}$	$\begin{align*}\csc x&=\dfrac{1}{\sin x}\\&=\dfrac{\sqrt{2}}{\sin a-\cos a}\end{align*}$	$\begin{align*}\sec x+\csc x&=-\dfrac{\sqrt{2}}{\sin a+\cos a}+\dfrac{\sqrt{2}}{\sin a-\cos a}\\&=\dfrac{2\sqrt{2}\cos a}{\sin^2 a-\cos^2 a}\end{align*}$

$\begin{align*}\therefore|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|&=|-\sqrt{2}\cos a+\dfrac{2}{\cos^2 a-\sin^2 a}+\dfrac{2\sqrt{2}\cos a}{\sin^2 a-\cos^2 a}|\\&=|-\sqrt{2}\cos a+\dfrac{2}{\cos^2 a-\sin^2 a}-\dfrac{2\sqrt{2}\cos a}{\cos^2 a-\sin^2 a}|\\&=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{\cos^2 a-\sin^2 a}|\\&=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{2\cos^2 a-1}|\\&=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{(\sqrt{2}\cos a-1)(\sqrt{2}\cos a+1)}|\\&=|-\sqrt{2}\cos a+\dfrac{2(1-\sqrt{2}\cos a)}{(\sqrt{2}\cos a-1)(\sqrt{2}\cos a+1)}|\\&=|-\sqrt{2}\cos a-\dfrac{2}{\sqrt{2}\cos a+1}|\\&=|-\left(\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \right)|\\&=\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1}\end{align*}$

If we have $(\sqrt{2}\cos a+1)$ and $\dfrac{2}{\sqrt{2}\cos a+1}$, 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

$(\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1}\ge 2 \sqrt{2}$

$\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \ge 2 \sqrt{2}-1$

We can conclude at this point the minimum value for \(\displaystyle |\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|\) is $2 \sqrt{2}-1$.

 

[Ackbach]

We can conclude at this point the minimum value for \(\displaystyle |\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|\) is $2 \sqrt{2}-1$.

You can only conclude that if there is a particular value of $a$ for which your function of $a$ actually achieves $2 \sqrt{2}-1$. Showing that $A \ge B$ does not imply $A=B$, or even that the minimum value of $A$ is $B$. Example: $x^{2}\ge -1$, but this does not mean the minimum value of $x^{2}$ is $-1$. For $x^{2} \ge -1$, we say the inequality is not tight. But if we have the inequality $x^{2} \ge 0$, then the inequality is tight, because you get equality for at least one value of $x$.

So: is your last inequality there tight or not, according to this line of reasoning?

 

[anemone]

You can only conclude that if there is a particular value of $a$ for which your function of $a$ actually achieves $2 \sqrt{2}-1$. Showing that $A \ge B$ does not imply $A=B$, or even that the minimum value of $A$ is $B$. Example: $x^{2}\ge -1$, but this does not mean the minimum value of $x^{2}$ is $-1$. For $x^{2} \ge -1$, we say the inequality is not tight. But if we have the inequality $x^{2} \ge 0$, then the inequality is tight, because you get equality for at least one value of $x$.

So: is your last inequality there tight or not, according to this line of reasoning?

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:

$\begin{align*}\sin x&=\sin (a-135^{\circ})\\&=\dfrac{-\sin a-\cos a}{\sqrt{2}}\end{align*}$	$\begin{align*}\cos x&=\cos (a-135^{\circ})\\&=\dfrac{-\cos a+\sin a}{\sqrt{2}}\end{align*}$	$\begin{align*}\sin x+\cos x&=\dfrac{-\sin a-\cos a}{\sqrt{2}}+\dfrac{-\cos a+\sin a}{\sqrt{2}}\\&=-\sqrt{2}\cos a\end{align*}$
$\begin{align*}\tan x&=\tan(a-135^{\circ})\\&=\dfrac{\sin a+\cos a}{\cos a-\sin a}\end{align*}$	$\begin{align*}\cot x&=\cot(a-135^{\circ})\\&=\dfrac{\cos a-\sin a}{\sin a+\cos a}\end{align*}$	$\begin{align*}\tan x+\cot x&=\dfrac{\sin a+\cos a}{\cos a-\sin a}+\dfrac{\cos a-\sin a}{\sin a+\cos a}\\&=\dfrac{2}{\cos^2 a-\sin^2 a}\end{align*}$
$\begin{align*}\sec x&=\dfrac{1}{\cos x}\\&=-\dfrac{\sqrt{2}}{\sin a+\cos a}\end{align*}$	$\begin{align*}\csc x&=\dfrac{1}{\sin x}\\&=\dfrac{\sqrt{2}}{\sin a-\cos a}\end{align*}$	$\begin{align*}\sec x+\csc x&=-\dfrac{\sqrt{2}}{\sin a+\cos a}+\dfrac{\sqrt{2}}{\sin a-\cos a}\\&=\dfrac{2\sqrt{2}\cos a}{\sin^2 a-\cos^2 a}\end{align*}$

$\begin{align*}\therefore|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|&=|-\sqrt{2}\cos a+\dfrac{2}{\cos^2 a-\sin^2 a}+\dfrac{2\sqrt{2}\cos a}{\sin^2 a-\cos^2 a}|\\&=|-\sqrt{2}\cos a+\dfrac{2}{\cos^2 a-\sin^2 a}-\dfrac{2\sqrt{2}\cos a}{\cos^2 a-\sin^2 a}|\\&=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{\cos^2 a-\sin^2 a}|\\&=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{2\cos^2 a-1}|\\&=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{(\sqrt{2}\cos a-1)(\sqrt{2}\cos a+1)}|\\&=|-\sqrt{2}\cos a+\dfrac{2(1-\sqrt{2}\cos a)}{(\sqrt{2}\cos a-1)(\sqrt{2}\cos a+1)}|\\&=|-\sqrt{2}\cos a-\dfrac{2}{\sqrt{2}\cos a+1}|\\&=|-\left(\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \right)|\\&=|\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1}|\end{align*}$

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 $(\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1}\ge 2 \sqrt{2}$ $\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \ge 2 \sqrt{2}-1$

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 $-\left((\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1} \right)\ge 2 \sqrt{2}$ $(\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1} \le -2 \sqrt{2}$ $\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \le -2 \sqrt{2}-1$

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 \(\displaystyle |\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|\) is $2 \sqrt{2}-1$.

Does this look okay now, Ackbach?

 

[Ackbach]

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:

$\begin{align*}\sin x&=\sin (a-135^{\circ})\\&=\dfrac{-\sin a-\cos a}{\sqrt{2}}\end{align*}$	$\begin{align*}\cos x&=\cos (a-135^{\circ})\\&=\dfrac{-\cos a+\sin a}{\sqrt{2}}\end{align*}$	$\begin{align*}\sin x+\cos x&=\dfrac{-\sin a-\cos a}{\sqrt{2}}+\dfrac{-\cos a+\sin a}{\sqrt{2}}\\&=-\sqrt{2}\cos a\end{align*}$
$\begin{align*}\tan x&=\tan(a-135^{\circ})\\&=\dfrac{\sin a+\cos a}{\cos a-\sin a}\end{align*}$	$\begin{align*}\cot x&=\cot(a-135^{\circ})\\&=\dfrac{\cos a-\sin a}{\sin a+\cos a}\end{align*}$	$\begin{align*}\tan x+\cot x&=\dfrac{\sin a+\cos a}{\cos a-\sin a}+\dfrac{\cos a-\sin a}{\sin a+\cos a}\\&=\dfrac{2}{\cos^2 a-\sin^2 a}\end{align*}$
$\begin{align*}\sec x&=\dfrac{1}{\cos x}\\&=-\dfrac{\sqrt{2}}{\sin a+\cos a}\end{align*}$	$\begin{align*}\csc x&=\dfrac{1}{\sin x}\\&=\dfrac{\sqrt{2}}{\sin a-\cos a}\end{align*}$	$\begin{align*}\sec x+\csc x&=-\dfrac{\sqrt{2}}{\sin a+\cos a}+\dfrac{\sqrt{2}}{\sin a-\cos a}\\&=\dfrac{2\sqrt{2}\cos a}{\sin^2 a-\cos^2 a}\end{align*}$

$\begin{align*}\therefore|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|&=|-\sqrt{2}\cos a+\dfrac{2}{\cos^2 a-\sin^2 a}+\dfrac{2\sqrt{2}\cos a}{\sin^2 a-\cos^2 a}|\\&=|-\sqrt{2}\cos a+\dfrac{2}{\cos^2 a-\sin^2 a}-\dfrac{2\sqrt{2}\cos a}{\cos^2 a-\sin^2 a}|\\&=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{\cos^2 a-\sin^2 a}|\\&=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{2\cos^2 a-1}|\\&=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{(\sqrt{2}\cos a-1)(\sqrt{2}\cos a+1)}|\\&=|-\sqrt{2}\cos a+\dfrac{2(1-\sqrt{2}\cos a)}{(\sqrt{2}\cos a-1)(\sqrt{2}\cos a+1)}|\\&=|-\sqrt{2}\cos a-\dfrac{2}{\sqrt{2}\cos a+1}|\\&=|-\left(\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \right)|\\&=|\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1}|\end{align*}$

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 $(\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1}\ge 2 \sqrt{2}$ $\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \ge 2 \sqrt{2}-1$

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 $-\left((\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1} \right)\ge 2 \sqrt{2}$ $(\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1} \le -2 \sqrt{2}$ $\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \le -2 \sqrt{2}-1$

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 \(\displaystyle |\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|\) is $2 \sqrt{2}-1$.

Does this look okay now, Ackbach?

Very excellent! You've convinced me that the function actually achieves that minimum, hence the inequality is tight.

Oh, and since you didn't really have to add all that much to finish it, I wouldn't have said your previous post was exceptionally incomplete.