Given that cos x = \(\frac{12}{13}\), evaluate \(\frac{1 - \tan x}{\tan x}\)

We can use trigonometric identities to evaluate the expression. We know that: - $\tan x = \frac{\sin x}{\cos x}$. - $\sin^2x + \cos^2x = 1$. We can use the second identity to find $\sin x$: $$\begin{aligned} \sin^2x + \cos^2x &= 1 \\ \sin^2x &= 1 - \cos^2x \\ \sin x &= \sqrt{1 - \cos^2x} \\ &= \sqrt{1 - \left(\frac{12}{13}\right)^2} \\ &= \frac{5}{13} \end{aligned}$$ Now we can substitute $\sin x$ and $\cos x$ into the expression we need to evaluate: $$\begin{aligned} \frac{1 - \tan x}{\tan x} &= \frac{1 - \frac{\sin x}{\cos x}}{\frac{\sin x}{\cos x}} \\ &= \frac{\cos x - \sin x}{\sin x} \\ &= \frac{\frac{12}{13} - \frac{5}{13}}{\frac{5}{13}} \\ &= \frac{7}{5} \end{aligned}$$ Therefore, the answer is $\frac{7}{5}$.