Find the minimum and maximum value of $\frac{1}{\sin{x} -3\cos{x} +5}$ Find the maximum value and minimum value of
$$\frac{1}{\sin{x} -3\cos{x} +5}.$$ Any tips?
First notice the following:

$\sin{x} -3\cos{x}$

$=\sqrt{10}(\frac{1}{\sqrt{10}}\sin{x} - \frac{3}{\sqrt{10}}\cos{x})$

$=\sqrt{10}(\cos{\alpha}\sin{x} +\sin{\alpha}\cos{x})$; (where $\tan{\alpha} = -3)$

$=\sqrt{10}\sin{(x+\alpha)}$

Now, since the 'sine' function lies between -1 and 1, the expression "$\sin{x} -3\cos{x}$" lies between $-\sqrt{10}$ and $\sqrt{10}$. Put these values for the epression in your overall expression to find it's range.
$-\sqrt {10}+5\leq \sin x-3\cos x+5\leq \sqrt {10}+5\\ \implies \frac{1}{\sqrt {10}+5}\leq \frac{1}{\sin x-3\cos x+5}\leq \frac{1}{-\sqrt {10}+5}$