You may approach the problem proving that `tan 2x = (2tan x)/(1 - tan^2 x) = 1/(1 - tan x) - 1/(1 + tan x` ) such that:

`(2tan x)/(1 - tan^2 x) = 2tan x/((1 - tan x)(1 + tan x))`

You need to use partial fraction decomposition, such that:

`2tan x/((1 - tan x)(1 + tan x)) = A/(1 - tan x) + B/(1 + tan x)`

`2tan x = A(1 + tan x) + B(1 - tan x)`

`2tan x = A + A(tan x) + B - Btan x`

`2 tan x = tan x(A - B) + A + B`

Equating the coefficient of like parts yields:

`{(A - B = 2),(A + B = 0)} => 2A = 2 => A = 1 => B = -1`

`2tan x/((1 - tan x)(1 + tan x)) = 1/(1 - tan x) - 1/(1 + tan x)`

**Hence, testing if `1/1 - tan x) - 1/(1 + tan x) = tan 2x` , using double angle identity and partial fraction decomposition, yields `1/(1 - tan x) - 1/(1 + tan x) = tan 2x.` **

To add the ratios, we'll have to have the same denominator. We'll calculate the least common denominator for the given ratios.

LCD = (1-tanx)(1+tanx)

We notice that the LCD is a difference of square:

LCD = 1 - (tan x)^2

To obtain the denominator 1 - (tan x)^2, we'll multiply the first ratio by (1+tanx) and the second ratio, by (1-tanx).

We'll re-write the left side:

1/(1-tanx) - 1/(1+tanx) = (1 + tan x - 1 + tan x)/[1 - (tan x)^2]

We'll eliminate and combine the like terms from numerator:

1/(1-tanx) - 1/(1+tanx) = 2tan x/[1 - (tan x)^2] (1)

Now, we'll re-write the right side:

tan 2x = tan (x+x)

We'll apply the formula for the tangent of the sum of 2 angles:

tan (x+x) = (tan x + tan x)/(1-tan x*tan x)

tan 2x = 2tan x/[1 - (tan x)^2] (2)

We notice that we have obtained (1) = (2), so the identity is verified, fro any value of x:

1/(1-tanx) - 1/(1+tanx) = 2 tan x