Question

If $\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$, then the value of $\frac{\tan x}{\tan y}$ equals

• A. $\frac{b}{a}$
• B. $\frac{a}{b}$
• C. $1$
• D. $0$

Answer 1

The correct option is B $\frac{a}{b}$

$\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$

Using componendo and dividendo

$\frac{\sin (x+y)+\sin (x-y)}{\sin (x+y)-\sin (x-y)}=\frac{a}{b}$

$\frac{2\sin \left(\frac{(x+y)+(x-y)}{2}\right)\cos \left(\frac{(x+y)-(x-y)}{2}\right)}{2\sin \left(\frac{(x+y)-(x-y)}{2}\right)\cos \left(\frac{(x+y)+(x-y)}{2}\right)}=\frac{a}{b}$

$\frac{\sin x.\cos y}{\sin y.\cos x}=\frac{a}{b}$

$\frac{\tan x}{\tan y}=\frac{a}{b}$