Let $\triangle ABC$ be a right-angled triangle, right-angled at point $B$.

Given that,

$\sin A=\frac{3}{4}$

$\frac{B C}{A C}=\frac{3}{4}$

Let $BC$ be $3 k$. Therefore, $AC$ will be $4 k,$ where $k$ is a positive integer.

Applying Pythagoras theorem in $\triangle ABC$, we obtain

$AC ^{2}= AB ^{2}+ BC ^{2}$

$(4 k)^{2}= AB ^{2}+(3 k)^{2}$

$16 k^{2}-9 k^{2}=A B^{2}$

$7 k^{2}=A B^{2}$

$A B=\sqrt{7} k$

$\cos A=\frac{\text { Side adjacent to } \angle A}{\text { Hypotenuse }}$

$=\frac{A B}{A C}=\frac{\sqrt{7 }k}{4 k}=\frac{\sqrt{7}}{4}$

$\tan A=\frac{\text { Side opposite to } \angle A}{\text { Side adjacent to } \angle A}$

$=\frac{B C}{A B}=\frac{3 k}{\sqrt{7} k}=\frac{3}{\sqrt{7}}$