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 $\frac{x}{y}$

$\tan A=\frac{x\sin B}{1-x\cos B}$

$\cot A=\frac{1-x\cos B}{x\sin B}=\frac{1}{x\sin B}-\frac{\cos B}{x\sin B}$

$\cot A=\frac{1}{x\sin B}-\cot B$

$\cot A+\cot B=\frac{1}{x\sin B}$ ……..$\left(1\right)$

$\tan B=\frac{y\sin A}{1-y\cos A}$

$\cot B=\frac{1-y\cos A}{y\sin A}=\frac{1}{y\sin A}-\cot A$

$\cot A+\cot B=\frac{1}{y\sin A}$ ……..$\left(2\right)$

from $\left(1\right)$ and $\left(2\right)$

$\frac{1}{x\sin B}=\frac{1}{y\sin A}$

$\frac{\sin A}{\sin B}=\frac{x}{y}$.