Answer
$\dfrac{-3}{4}$
Work Step by Step
Suppose $\theta =\sin^{-1} (\dfrac{-3}{5})$
This gives: $\sin \theta=-\dfrac{3}{5}$
Since, $ r=\sqrt {x^2+y^2}$
This implies that $ x=\sqrt {r^2-y^2}=\sqrt {5^2-3^2}=4$
Therefore, we have $\tan \theta =\tan [\sin^{-1} (\dfrac{-3}{5})]=\dfrac{-3}{4}$
