If sin x

We have:

$\sin x=\frac{12}{13}$ and $x$ lie in the second quadrant.

In the second quadrant, $\sin x$ and $\operatorname{cosec} x$ are positive and all the other four $\mathrm{T}-$ ratios are negative.

$\therefore \cos x=-\sqrt{1-\sin ^{2} x}$

$=-\sqrt{1-\left(\frac{12}{13}\right)^{2}}$

$=\frac{-5}{13}$

$\tan x=\frac{\sin x}{\cos x}$

$=\frac{\frac{12}{13}}{\frac{-5}{13}}$

$=\frac{-12}{5}$

And, $\sec x=\frac{1}{\cos x}$

$=\frac{1}{\frac{-5}{13}}$

$=\frac{-13}{5}$

$\therefore \sec x+\tan x=\frac{-13}{5}+\frac{-12}{5}$

$=-5$