since $\cot x=-\frac{5}{12},$ we have $\tan x=-\frac{12}{5}$

Now      $\sec ^{2} x=1+\tan ^{2} x=1+\frac{144}{25}=\frac{169}{25}$

Hence $\sec x=\pm \frac{13}{5}$

since $x$ lies in second quadrant, sec $x$ will be negative. Therefore

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

which also gives

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

Further, we have

$\sin x =\tan x \cos x=\left(-\frac{12}{5}\right) \times\left(-\frac{5}{13}\right)=\frac{12}{13} $

and   $\cos ec\, x =\frac{1}{\sin x}=\frac{13}{12}$