Correct option is D

Given:

​$$\frac{\sin(x - y)}{\sin(x + y)} \cdot \frac{\tan x + \tan y}{\tan x - \tan y}$$​​

Formula Used:

We will use the following trigonometric identities:

​$$\sin(A - B) = \sin A \cos B - \cos A \sin B\\\ \\\sin(A + B) = \sin A \cos B + \cos A \sin B\\\ \\\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\\\ \\.\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$​​

Solution:

First, express sin(x - y) and sin(x + y) using their identities:

​$$\sin(x - y) = \sin x \cos y - \cos x \sin y\\\ \\\sin(x + y) = \sin x \cos y + \cos x \sin y$$​​

Next, simplify the expression using the given values for \tan x and \tan y:

​$$\frac{\sin(x - y)}{\sin(x + y)} \cdot \frac{\tan x + \tan y}{\tan x - \tan y}$$ 

​$$\frac{\sin(x - y)}{\sin(x + y)} \times \frac{\tan x + \tan y}{\tan x - \tan y}\\\ \\\sin(x - y) = \sin(x)\cos(y) - \cos(x)\sin(y), \quad \sin(x + y) = \sin(x)\cos(y) + \cos(x)\sin(y), \\\ \\\tan(x) = \frac{\sin(x)}{\cos(x)}, \quad \tan(y) = \frac{\sin(y)}{\cos(y)}. \\\ \\\frac{\sin x \times \cos y - \cos x \times \sin y}{\sin x \times \cos y + \cos x \times \sin y} \times \frac{\sin x \times \cos y + \sin y \times \cos x}{\sin x \times \cos y - \sin y \times \cos x}.$$ 

= 1 

​​After simplification, the final value of the given expression is 1.

Alternative Solution: