<s>[INST] Find the minimum value of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ for real numbers $x$ I'm tutoring for a college math class and we're doing putnam problems next week and this one stumped me: 

Find the minimum value of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ for real numbers $x$.

 [/INST] Let $x \in [0,2 \pi)$ and
$$f(x) = \sin{x} + \cos{x}+\tan{x}+\cot{x}+\sec{x}+\csc{x}$$
Then
$$\begin{align}f'(x) &= \cos{x} - \sin{x} + \sec^2{x} - \csc^2{x} + \sec{x} \tan{x} - \csc{x} \cot{x}\\ &= \frac{\cos{x}-\sin{x}}{\cos^2{x} \sin^2{x}} (\cos^2{x} \sin^2{x} - (\cos{x} + \sin{x}) - (\cos^2{x}+\cos{x} \sin{x} + \sin^2{x}))\\ &= \frac{\cos{x}-\sin{x}}{\cos^2{x} \sin^2{x}} (\cos{x} \sin{x} - \cos{x} - \sin{x})(1 + \cos{x} \sin{x} + \cos{x}+\sin{x}) \end{align}$$
Regardless of the absolute value, we have extrema where
$$\begin{array} \\ \cos{x} - \sin{x} = 0 & (1) \\ \cos{x} \sin{x} - \cos{x} - \sin{x}=0 & (2) \\ 1 + \cos{x} \sin{x} + \cos{x}+\sin{x}=0 & (3) \end{array}$$
(1) implies that $x \in \{\pi/4,5 \pi/4\}$, while (3) implies that $x \in \{\pi,3 \pi/2\}$.  (2) on the other hand implies that
$$\frac{1}{4} \sin^2{2 x} = 1 + \sin{2 x} $$
or that $2 x = 2 \pi-\arcsin{[2 (\sqrt{2}-1)]}$ so that the solution is $\in [0,2\pi)$.  (The other solution to the quadratic is $> 1$.)
Plugging in these values into $g(x)$, we find that
$$\begin{align}\left |g \left ( \frac{\pi}{4} \right ) \right | &= 3 \sqrt{2} + 2\\ \left |g \left ( \frac{5\pi}{4} \right ) \right | &= 3 \sqrt{2} - 2\\ |g(\pi)| &= 2\\ \left |g \left ( \frac{3\pi}{2} \right ) \right | &=2\\\end{align}$$
Finally, for the last solution, I will spare you the simplifications involved; we get
$$\left|g \left ( \pi - \frac{1}{2} \arcsin{[2 (\sqrt{2}-1)]} \right ) \right | = 2 \sqrt{2}-1 $$
The minimum value is thus $2 \sqrt{2}-1 $.
EDIT
Per the point raised by @Ivan below, I show that $f(x) \ne 0 \: \forall x \in [0,2 \pi)$.  $f(x)$ may be rewritten as 
$$f(x) = \frac{\csc \left(\frac{\pi }{4}-\frac{x}{2}\right) \csc \left(\frac{x}{2}\right)
   (-\sin (x)-\cos (x)+2 \sin (x) \cos (x)+3)}{2 \sqrt{2}}$$
The solution to $f(x)=0$ means that
$$\sin^2{2 x} + 5 \sin{2 x} + 8 = 0$$
which has no real solutions.  Thus, $f(x) \ne 0 \: \forall x \in [0,2 \pi)$ and the result holds.
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