A geometry problem by Rakshit Joshi

$\begin{aligned} \frac{\color{#3D99F6}{\sin (2x)}}{\color{#D61F06}{\sin (2y)}} & = \color{#3D99F6}{\frac{2\tan x}{1+\tan^2 x}} \times \color{#D61F06}{\frac{1+\tan^2 y}{2\tan y}} \quad \quad \small \text{Using half-angle identity} \\ & = \frac{\tan x (1+\tan^2 y)}{\tan y (1+\tan^2 x)} \end{aligned}$

Now evaluating:

$\begin{aligned} (a+b)\tan (x-y) & = (a-b)\tan(x+y) \\ \frac{(a+b)(\tan x - \tan y)}{1+\tan x \tan y} & = \frac{(a-b)(\tan x + \tan y)}{1-\tan x \tan y} \\ (a+b)(\tan x - \tan y)(1-\tan x \tan y) & = (a-b)(\tan x + \tan y)(1 + \tan x \tan y) \\ (a+b)(\tan x - \tan y -\tan^2 x \tan y + \tan x \tan^2 y) & = (a-b)(\tan x + \tan y +\tan^2 x \tan y + \tan x \tan^2 y) \\ b(\tan x + \tan x \tan^2 y) & = a(\tan y + \tan^2 x \tan y) \\ \frac{\tan x (1+\tan^2 y)}{\tan y (1+\tan^2 x)} & = \frac{a}{b} \\ \implies \frac{\sin (2x)}{\sin (2y)} & = \boxed{\dfrac{a}{b}} \end{aligned}$

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

Applying componendo-dividendo,
$\dfrac{a}{b} = \dfrac{\tan(x+y)+ \tan(x-y)}{\tan(x+y) - \tan(x-y)}$
$\dfrac{a}{b} = \dfrac{\dfrac{\sin(x+y)\cos(x-y) + \cos(x+y)\sin(x-y)}{\cos(x+y)\cos(x-y)}}{\dfrac{\sin(x+y)\cos(x-y) - \cos(x+y)\sin(x-y)}{\cos(x+y)\cos(x-y)}}$
$\dfrac{a}{b} = \dfrac{\sin(x+y+x-y)}{\sin(x+y-(x-y))} = \dfrac{\sin(2x)}{\sin(2y)}$