Answer
$\dfrac{12}{5}$
Work Step by Step
Suppose $\theta =\sin^{-1} (\dfrac{5}{13})$
This gives: $\sin \theta=\dfrac{5}{13}$
Since, $ r=\sqrt {x^2+y^2}$
This implies that $ x=\sqrt {r^2-y^2}=\sqrt {13^2-5^2}=12$
Therefore, we have $\cot\theta =\cot [\sin^{-1} (\dfrac{5}{13})]=\dfrac{12}{5}$
