Solution 1:
$$\dfrac{\dfrac{\sin{A} + \cos{A}}{\sin{A} \cos{A}}}{\dfrac{\sin^2{A} + \cos^2{A}}{\sin{A} \cos{A}}}$$
$$ = \frac{\sin{A} + \cos{A}}{\sin^2{A} + \cos^2{A}}$$
$$ = \sin{A} + \cos{A}$$
Solution 2:
$$\frac{\cos{A}(1 + \cos{A}) + \sin^2{A}}{\sin{A} (1 + \cos{A})}$$
$$= \frac{\color{red}{\cos{A} + 1}}{\sin{A} (\color{red}{\cos{A} + 1})}$$
$$= \frac{1}{\sin{A}} = \csc{A}$$
PS: I don't know how to put those cross-marks(cancellations) on fractions, if someone knows, please comment it, I'll edit it.
Let $\sin{x}+\cos{x}=\sqrt{2}\sin{(x+\frac{\pi}{4})}=a$, then $a$ can take any value between $-\sqrt{2}$ and $\sqrt{2}$. We have $\sin{x}\cos{x}=\frac{a^2-1}{2}$. Thus
\begin{align}
\left||\sin{x}+\cos{x}+\tan{x}+\cot{x}+\sec{x}+\csc{x} \right|& =\left|\sin{x}+\cos{x}+\frac{1}{\sin{x}\cos{x}}+\frac{\sin{x}+\cos{x}}{\sin{x}\cos{x}} \right| \\
& =\left|a+\frac{2+2a}{a^2-1} \right| \\
& =\left|a+\frac{2}{a-1} \right|
\end{align}
If $a<1$, let $b=1-a>0$, then $a+\frac{2}{a-1}=1-(b+\frac{2}{b}) \leq 1-2\sqrt{2}<0$, so $\left|a+\frac{2}{a-1} \right| \geq 2\sqrt{2}-1$.
($b+\frac{2}{b} \geq 2\sqrt{2}$ by AM-GM inequality or square $\geq 0$)
Equality holds when $b=\sqrt{2}, a=1-\sqrt{2}, x=\sin^{-1}{(\frac{1-\sqrt{2}}{\sqrt{2}})}-\frac{\pi}{4}$.
If $1 \leq a \leq \sqrt{2}$, then $a+\frac{2}{a-1}>\frac{2}{a-1}>2$, so $\left|a+\frac{2}{a-1} \right|>2>2\sqrt{2}-1$.
Thus the minimum value is $2\sqrt{2}-1$, achieved when $x=\sin^{-1}{(\frac{1-\sqrt{2}}{\sqrt{2}})}-\frac{\pi}{4}$.
Best Answer
Please have a look at this: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2325342&sid=03b52017fa0ef5480a4573048ae117c1#p2325342