Question

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

• A. $\frac{a^2}{b^2}$
• B. $\frac{a}{b}$
• C. $\frac{b}{a}$
• D. $\frac{\left(a^2+b^2\right)}{\left(a^2-b^2\right)}$

Answer 1

The correct option is B

$\frac{a}{b}$

Explanation for the correct answer

Given information

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

Using componendo and dividendo rule

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

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

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

$\Rightarrow \tan x.coty=\frac{a}{b}$

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

Hence option (B) is the correct option.