<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] Let the value of this expression at $x=p$ be $q$. Then we have
$${3+\cos p\over\sin p}= q$$
$$\implies\cos p-q\sin p = -3$$
Dividing both LHS and RHS by $\sqrt{q^2+1}$ we have 
$$\cos p\cdot {1\over\sqrt{q^2+1}} - \sin p\cdot{q\over\sqrt{q^2+1}} = {-3\over\sqrt{q^2+1}}$$
Let $1\over\sqrt{q^2+1}$ be $\cos r$. So, we have $$\cos p\cos r - \sin p\sin r = {-3\over\sqrt{q^2+1}}$$
Or $$\cos{(p+r)} = \frac{-3}{\sqrt{q^2+1}}$$
For this to be a valid expression $\sqrt{q^2+1}$ must be greater than $3$ since $\cos x \in [-1,1]$. So, we have 
$$q^2+1 \geq 9$$
$$\implies |q| \geq \sqrt 8$$
$$\implies q \in \left(-\infty,-2\sqrt 2\right]\cup\left[2\sqrt 2, \infty\right)$$
$$\implies\boxed{ {3+\cos x\over\sin x}\notin\left(-2\sqrt 2, 2\sqrt 2\right)}$$
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