Find the values of the other five trigonometric functions if $\sec x = \frac{13}{5}$ and $x$ lies in the fourth quadrant.

(N/A) Given $\sec x = \frac{13}{5}$.
Since $\cos x = \frac{1}{\sec x}$,we have $\cos x = \frac{5}{13}$.
Using the identity $\sin^2 x + \cos^2 x = 1$,we get $\sin^2 x = 1 - \cos^2 x = 1 - (\frac{5}{13})^2 = 1 - \frac{25}{169} = \frac{144}{169}$.
Thus,$\sin x = \pm \frac{12}{13}$.
Since $x$ lies in the $4^{\text{th}}$ quadrant,$\sin x$ is negative,so $\sin x = -\frac{12}{13}$.
Now,$\csc x = \frac{1}{\sin x} = -\frac{13}{12}$.
$\tan x = \frac{\sin x}{\cos x} = \frac{-12/13}{5/13} = -\frac{12}{5}$.
$\cot x = \frac{1}{\tan x} = -\frac{5}{12}$.