If tan(x + y) tan(x - y) = 1, then the value of tan x is :

$$tan(x + y) = \frac{tan x + tan y}{1- tan x tan y}$$

$$tan(x - y) = \frac{tan x - tan y}{1+ tan x tan y}$$

$$tan(x+y)tan(x-y) =1$$

$$ \frac{tan x + tan y}{1- tan x tan y} \times \frac{tan x - tan y}{1+ tan x tan y} =1 $$

$$ \frac{tan^{2} x - tan^{2} y}{1 - tan^{2} x tan ^{2} y} =1$$

$$ tan^{2} x - tan^{2} y =1 - tan^{2} x tan ^{2} y$$

$$ tan^{2} x + tan^{2} x tan^{2} y =1 + tan ^{2} y$$

$$ tan^{2} x (1 + tan^{2} y) =1 + tan ^{2} y$$

$$ tan^{2} x =1$$

$$ tan x =1$$

Option B is the correct answer.