Question

If $\sin x\cos y=\frac{1}{4}$ and $3\tan x=4\tan y,$ , then $\sin (x-y)$ equals to

• A. $\frac{1}{16}$
• B. $\frac{1}{8}$
• C. $\frac{3}{16}$
• D. $\frac{3}{4}$

Answer 1

Answer: (A)

Given, $\sin x\cos y=\frac{1}{4}$ and $3\tan x=4\tan y$
$\therefore$ $\frac{3\sin x}{\cos x}=\frac{4\sin y}{\cos y}$
$\Rightarrow$ $3\sin x\cos y=4\sin y\cos x$
$\Rightarrow$ $3\times \frac{1}{4}=4\sin y\cos x$
$\Rightarrow$ $\sin y\cos x=\frac{3}{16}$
$\therefore$ $\sin (x-y)=\sin x\cos y-\cos x\sin y$
$=\frac{1}{4}-\frac{3}{16}$ $=\frac{1}{16}$