Answer
$\dfrac{-13}{12}$
Work Step by Step
Apply the trigonometric identities:
$\sin^2 \theta+\cos^2 \theta=1 \implies \sin \theta=-\sqrt{1-\cos^2\theta}$
and $\sec^2 \theta=1+\tan^2 \theta$
Since, the $\tan \theta$ is positive and $\sin \theta$ is negative in second quadrant, so does $\sec \theta$ is negative.
Thus, we have
$\cos \theta=-\sqrt{1-({\dfrac{5}{13})^2}}=\dfrac{-12}{13}$
Also, $\sec \theta =\dfrac{1}{\cos \theta}=\dfrac{1}{\dfrac{-12}{13}}=\dfrac{-13}{12}$
