Question

If $\frac{sin\left(x+y\right)}{sin\left(x-y\right)}=\frac{a+b}{a-b}$ , then $\frac{tanx}{tany}$ 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
$\frac{\sin (x+y)+\sin (x-y)}{\sin (x+y)-\sin (x-y)}=\frac{(a+b)+(a-b)}{(a+b)-(a-b)}$
$\Rightarrow \frac{2\sin x\cos y}{2\cos x\sin y}=\frac{2a}{2b}$
$\Rightarrow \frac{\tan x}{\tan y}=\frac{a}{b}$