help

STEP 1.

So, we have a right triangle $\triangle {ABC}$ with $\angle B = 90^\circ$.

Let sides ${AC} = a$, $AB = b$. and $CB = c$.

$2\sin A$ is equal to $2(\frac{c}{a})$, and $3\cos A$ is equal to $3(\frac{b}{a})$

Therfore, 

 $2(\frac{c}{a}) = 3(\frac{b}{a})$,

$\frac{2c}{a} = \frac{3b}{a}$,

$2c = 3b$.

STEP 2.

$\tan A$ is equal to $\frac{c}{b}$.

We can substitute $c$ for $\frac{3b}{2}$.

$\tan A$ $=$ $\frac{\frac{3b}{2}}{b}$,

$\tan A$ $=$ $\frac{3b}{2} \times \frac{1}{b}$,

$\tan A$ $=$ $\boxed{\frac{3}{2}}$