Question

If $\tan A=\frac{x\sin B}{1-x\cos B}$ and $\tan B=\frac{y\sin A}{1-y\cos A}$, then the value of $\frac{\sin A}{\sin B}$ for all permissible values of $A,B$ is

• A. $\frac{x}{y}$
• B. $\frac{x+y}{x-y}$
• C. $\frac{y}{x}$
• D. $\frac{y+x}{y-x}$

Answer 1

The correct option is A $\frac{x}{y}$

$\tan A=\frac{x\sin B}{1-x\cos B}\Rightarrow \frac{\sin A}{\cos A}=\frac{x\sin B}{1-x\cos B}\Rightarrow \sin A-x\cos B\sin A=x\sin B\cos A\Rightarrow \sin A=x(\cos A\sin B+\cos B\sin A)\Rightarrow \sin A=x\sin (A+B)\cdots \left(1\right)$
$\tan B=\frac{y\sin A}{1-y\cos A}\Rightarrow \frac{\sin B}{\cos B}=\frac{y\sin A}{1-y\cos A}\Rightarrow \sin B-y\cos A\sin B=y\cos B\sin B\Rightarrow \sin B=y(\cos A\sin B+\cos B\sin B)\Rightarrow \sin B=y\sin (A+B)\cdots \left(2\right)$
Now,
$\frac{\sin A}{\sin B}=\frac{x\sin (A+B)}{y\sin (A+B)}\therefore \frac{\sin A}{\sin B}=\frac{x}{y}$