Question

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

• A. $\frac{a^2}{b^2}$
• B. $\frac{a}{b}$
• C. $\frac{b}{a}$
• D. $\frac{a^2+b^2}{a^2-b^2}$
• E. $\frac{a^2-b^2}{a^2+b^2}$

Answer 1

Answer: (B)

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