Sins

In the interval $\left[0 , 2 \pi \right)$ , we have

$\cos A < 0 \implies A \in \left( \dfrac{\pi}{2} , \dfrac{3 \pi}{2} \right)$

Here

$\sin A = \dfrac 35 > 0 \implies A \in \left( 0 , \pi \right)$

Thus, the common interval in which $A$ lies corresponds to $A \in \left( \dfrac{\pi}{2} , \pi \right)$ .

In this interval $\tan A < 0$ and so we have

$| \tan A | = \dfrac{|\sin A|}{|\cos A|} = \dfrac{\sin A}{\sqrt{1- \sin^2 A}} = \dfrac{{3}/{5}}{{4}/{5}} = \dfrac 34 \\ \implies \tan A = - \dfrac 34 = \boxed{- 0.75}$