If sin x cos y

Given $\sin x \cos y=\frac{1}{4}$

and $3 \tan x=4 \tan y$ i. e. $\tan x=\frac{4}{3} \tan y$

i. e. $\frac{\tan x}{\tan y}=\frac{4}{3}$

i. e. $\frac{\sin x}{\cos x} \frac{\cos y}{\sin y}=\frac{4}{3}$

$\Rightarrow \frac{1 / 4}{\cos x \sin y}=\frac{4}{3}$

$\Rightarrow \cos x \sin y=\frac{3}{16}$

$\therefore \sin (x-y)=\sin x \cos y-\cos x \sin y$

$=\frac{1}{4}-\frac{3}{16}$

$=\frac{4-3}{16}$

$\sin (x-y)=\frac{1}{16}$

hence, value of $\sin (x-y)=\frac{1}{16}$.