Question

If
$\rm \dfrac{\sin (x+y)}{\sin (x-y)}=\dfrac{a+b}{a-b}$
then find,
$\rm \dfrac{\tan x }{\tan y}$
​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\rm{\dfrac{sin(x+y)}{sin(x-y)}=\dfrac{a+b}{a-b}}$

$\rm{\implies\,\dfrac{sin(x)\,cos(y)+cos(x)\,sin(y)}{sin(x)\,cos(y)-cos(x)\,sin(y)}=\dfrac{a+b}{a-b}}$

Using componendo and dividendo, we get,

$\rm{\implies\,\dfrac{sin(x)\,cos(y)+cos(x)\,sin(y)+sin(x)\,cos(y)-cos(x)\,sin(y)}{sin(x)\,cos(y)+cos(x)\,sin(y)-sin(x)\,cos(y)+cos(x)\,sin(y)}=\dfrac{a+b+a-b}{a+b-a+b}}$

$\rm{\implies\,\dfrac{2\,sin(x)\,cos(y)}{2\,cos(x)\,sin(y)}=\dfrac{2\,a}{2\,b}}$

$\rm{\implies\,\dfrac{\dfrac{sin(x)}{cos(x)}}{\dfrac{sin(y)}{cos(y)}}=\dfrac{a}{b}}$

$\rm{\implies\,\dfrac{tan(x)}{tan(y)}=\dfrac{a}{b}}$