Question

If $\tan A=\frac{x\sin B}{1-x\cos B}$ and $\tan B=\frac{y\sin A}{1-y\cos A}$ then $\frac{\sin A}{\sin B}=$

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

Answer 1

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

$\frac{\sin A}{\cos A}=\frac{x\sin B}{1-x\cos B}\Rightarrow \sin A-x\sin A\cos B=x\cos A\sin B\Rightarrow \sin A=x\sin (A+B)$
Similarly $\sin B=y\sin (A+B).\therefore \frac{\sin A}{\sin B}=\frac{x}{y}.$