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} $