<s>[INST] Show that $\frac{3\> + \>\cos x}{\sin x}$ cannot have any value between $-2\sqrt2$ and $2\sqrt2$ Show that 
$$\dfrac{3+\cos x}{\sin x}\quad \forall \quad x\in R $$
cannot have any value between $-2\sqrt{2}$ and $2\sqrt{2}$.

My attempt is as follows:
There can be four cases, either $x$ lies in the first quadrant, second, third or fourth:-
First quadrant: $\cos x$ will decrease sharply and sinx will increase sharply, so $y_{min}=3$ at $x=\dfrac{\pi}{2}$.
$y_{max}$ would tend to $\infty$ near to $x=0$
Second quadrant: $\cos x$ will increase in magnitude and sinx will decrease sharply, so $y_{min}=3$ at $x=\dfrac{\pi}{2}$.
$y_{max}$ would tend to $\infty$ near to $x=\pi$
Third quadrant: $\cos x$ will decrease in magnitude and sinx will increase in magnitude but negative, so $y_{min}$ would tend to $-\infty$ near to $x=\pi$
$y_{max}$ would be $-3$ at $x=\dfrac{3\pi}{2}$
Fourth quadrant: $\cos x$ will increase sharply and sinx will decrease in magnitude, so $y_{min}$ would tend to $-\infty$ near to $x=2\pi$
$y_{max}$ would be $-3$ at $x=\dfrac{3\pi}{2}$
So in this way I have proved that $\dfrac{3+\cos x}{\sin x}$ cannot lie between $-2\sqrt{2}$ and $2\sqrt{2}$, but is their any smart solution so that we can calculate quickly.
 [/INST] Use the half-angle expressions 
$\cos x = \frac{1-\tan^2\frac x2}{1+\tan^2\frac x2}$ and 
$\sin x = \frac{2\tan\frac x2}{1+\tan^2\frac x2}$ to express 
$$I=\frac{3+\cos x}{\sin x}=
\frac{2}{\tan\frac x2} +\tan \frac x2$$
Note 
$$I^2=\left(\frac{2}{\tan \frac x2} +\tan \frac x2\right)^2
=\left(\frac{2}{\tan \frac x2} -\tan \frac x2\right)^2+8 \ge 8$$
Thus, $I^2$ can not have values within $[0,8)$, which means that $I=\frac{3+\cos x}{\sin x}$ can not have values within $(-2\sqrt2, \>2\sqrt2)$.
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