Question

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

• A. $ \frac{b}{a} $
• B. $ \frac{a}{b} $
• C. $ ab $
• D. None of these

Answer 1

Answer: (B)

Since, $\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$
Applying componendo and dividendo, we get
$\frac{\sin (x+y)+\sin (x-y)}{\sin (x+y)-\sin (x-y)}=\frac{(a+b)+(a-b)}{(a+b)-(a-b)}$
$\Rightarrow \frac{2 \sin x \cos y}{2 \cos x \sin y}=\frac{2 a}{2 b}$
$\Rightarrow \frac{\tan x}{\tan y}=\frac{a}{b}$