Prove that 1/(1-tanx) - 1/(1+tanx) = tan2x .

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