Answer
$\sec\theta=-\frac{13}{5},\sin\theta=-\frac{12}{13},\csc\theta=-\frac{13}{12},\tan\theta=\frac{12}{5},\cot\theta=\frac{5}{12}$
Work Step by Step
Since $\cos \theta=-\frac{5}{13}$,
we have $\sec \theta=-\frac{13}{5}$.
Now $\sin^{2}\theta+\cos^{2}\theta=1$, that is, $\sin^{2}\theta=1-\cos^{2}\theta$
or $\sin^{2}\theta=1-\frac{25}{169}=\frac{144}{169}$
Hence $\sin \theta= -\frac{12}{13}$ (As $\theta$ lies in QIII)
which gives $\csc \theta=-\frac{13}{12}$
Further, we have
$\tan \theta=\frac{\sin\theta}{\cos\theta}=\frac{12}{5}$ and
$\cot\theta=\frac{1}{\tan\theta}=\frac{5}{12}$
