If cos x

We have:

$\cos x=-\frac{3}{5}$ and $\pi

Thus, $x$ is in the third quadrant.

In the third quadrant, $\tan x$ and $\cot x$ are positive And, all the other four $\mathrm{T}$ - ratios are negative.

$\therefore \sin x=-\sqrt{1-\cos ^{2} x}=-\sqrt{1-\left(\frac{-3}{5}\right)^{2}}=\frac{-4}{5}$

$\tan x=\frac{\sin x}{\cos x}=\frac{\frac{-4}{5}}{-\frac{3}{5}}=\frac{4}{3}$

$\cot x=\frac{1}{\tan x}=\frac{1}{4 / 3}=\frac{3}{4}$

$\sec x=\frac{1}{\cos x}=\frac{1}{-3 / 5}=\frac{-5}{3}$

Now, $\frac{\cos e c \theta+\cot \theta}{\sec \theta-\tan \theta}=\frac{\frac{-5}{4}+\frac{3}{4}}{\frac{-5}{3}-\frac{4}{3}}$

$=\frac{-2 / 4}{-9 / 3}$

$=\frac{-1 / 2}{-3}=\frac{1}{6}$