Answer
$\dfrac{3}{\sqrt {13}}$ or, $\dfrac{3 \sqrt {13}}{13}$
Work Step by Step
Suppose $\theta =\tan^{-1} (-\dfrac{2}{3})$
This gives: $\tan \theta=(-\dfrac{2}{3})$
Since, $ r=\sqrt {x^2+y^2}$
This implies that $ r=\sqrt {3^2+2^2}=\sqrt {13}$
Therefore, we have $\cos \theta =\cos [\tan^{-1} (\dfrac{2}{3})]=-\dfrac{3}{\sqrt {13}}=\dfrac{3 \sqrt {13}}{13}$
We know that the answer is positive because we are in quadrant IV (since tan is negative).
