Find the values of the other five trigonometric functions if $\tan x = -\frac{5}{12}$ and $x$ lies in the second quadrant.

Given $\tan x = -\frac{5}{12}$.
$\cot x = \frac{1}{\tan x} = \frac{1}{(-\frac{5}{12})} = -\frac{12}{5}$.
Using the identity $1 + \tan^2 x = \sec^2 x$:
$1 + (-\frac{5}{12})^2 = \sec^2 x$
$1 + \frac{25}{144} = \sec^2 x$
$\frac{169}{144} = \sec^2 x$
$\sec x = \pm \frac{13}{12}$.
Since $x$ lies in the $2^{\text{nd}}$ quadrant,$\sec x$ is negative.
$\therefore \sec x = -\frac{13}{12}$.
$\cos x = \frac{1}{\sec x} = \frac{1}{(-\frac{13}{12})} = -\frac{12}{13}$.
Since $\tan x = \frac{\sin x}{\cos x}$,we have:
$\sin x = \tan x \times \cos x = (-\frac{5}{12}) \times (-\frac{12}{13}) = \frac{5}{13}$.
$\csc x = \frac{1}{\sin x} = \frac{1}{(\frac{5}{13})} = \frac{13}{5}$.