How to find the minimum of $f(x)=(\sin(x)+\cos(x)+\tan(x)+\cot(x)+\sec(x)+\csc(x))^2$? I need to find the minimum of $f(x)$ with 
$$f(x)=(\sin(x)+\cos(x)+\tan(x)+\cot(x)+\sec(x)+\csc(x))^2$$
Could you help me with some clues?
 A: You can do the following:
Let $$\sin x + \cos x = y$$
Then we have that $$\tan x + \cot x = \frac{2}{y^2 -1}$$
and
$$\sec x  + \csc x = \frac{2y}{y^2-1}$$
I believe $$\sin x + \cos x + \tan x + \cot x + \sec x  + \csc x  $$ simplifies to $$y + \frac{2}{y-1}$$ (but I tried doing it in my head, so might have made mistakes).
And you can use the standard calculus techniques, now. (but take care to eliminate the corner cases etc).
A: To built upon @Queayiouer if you split the function as
$$ f(x) = \left( g(x) \right)^2 =  \left( \left(\sin x+\frac{1}{\sin x}\right) + \left(\cos x+\frac{1}{\cos x}\right) + \left(\tan x+\frac{1}{\tan x}\right) \right)^2 $$
where the minimum occurs if $g(x)=0$ or $g'(x)=0$. The first is not going to happen.
So now we have 
$$ g(x) = g_1(x) + g_2(x) + g_3(x) = \left(\sin x+\frac{1}{\sin x}\right) + \left(\cos x+\frac{1}{\cos x}\right) + \left(\tan x+\frac{1}{\tan x}\right)  $$
Let find if they three functions have a common minimum since
$$ g_1'(x) = \cos(x) \left(1-\frac{1}{\sin^2 x} \right) = 0 $$
  $$ g_2'(x) = \sin(x) \left(\frac{1}{\cos^2 x}-1 \right) = 0 $$
  $$ g_3'(x) = \tan^2 x - \frac{1}{\tan^2 x} = 0$$
Unfortunately those are never all zero at the same time. So my guess it to proceed numerically with Newton-Raphson method where
$$ x \rightarrow x - \frac{ g_1(x) + g_2(x) + g_3(x) }{ g_1'(x) + g_2'(x) + g_3'(x) } $$
starting from $x=1$ or something. I get between $2.5<x<2.7$.
