Question

If $\frac{\sin (A+B)}{\sin (A-B)}=\frac{x+y}{x-y}$, then $\frac{\tan A}{\tan B}$ is equal to

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

Answer 1

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

$\frac{\sin (A+B)}{\sin (A-B)}=\frac{x+y}{x-y}$

Applying componendo and dividendo
$\Rightarrow \frac{\sin (A+B)+\sin (A-B)}{\sin (A+B)-\sin (A-B)}=\frac{x+y+(x-y)}{x+y-(x-y)}$
$\Rightarrow \frac{2\sin A\cos B}{2\cos A\sin B}=\frac{2x}{2y}$
$\Rightarrow \frac{\tan A}{\tan B}=\frac{x}{y}$