Demonstrate that cos2x(1+tanx*tan2x)=1

The first step will be to divide both sides by cos 2x.
This division is possible because cos 2x is not cancelling (if cos 2x is cancelling, the
product cos2x(1+tanx*tan2x) would be 0 and not
1).

1+tanx*tan2x = 1 / cos
2x

Now, we'll substitute tan 2x by tan
(x+x):

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

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

1 + tanx*tan2x = 1 + tan x*{2tan x/[1 - (tan
x)^2]}

We'll multiply 1 by [1 - (tan x)^2] and we'll
get:

1 + tanx*tan2x = [1 - (tan x)^2 + 2(tan x)^2]/[1 -
(tan x)^2]

We'll combine like
terms:

1 + tanx*tan2x = [1 + (tan x)^2]/[1 - (tan
x)^2]

We'll write tan x = sin x/cos
x

We'll square raise both
sides:

(tan x)^2 = (sin x)^2/(cos
x)^2

We'll write the
numerator:

1 + (tan x)^2 = 1 +  (sin x)^2/(cos
x)^2

1 +  (sin x)^2/(cos x)^2 = [(cos x)^2 + (sin
x)^2]/(cos x)^2

From the fundamental formula of
trigonometry, we'll have:

(cos x)^2 + (sin x)^2 =
1

1 + (tan x)^2 = 1/(cos x)^2
(1)

1 - (tan x)^2 = 1 - (sin x)^2/(cos
x)^2

1 - (sin x)^2/(cos x)^2 = [(cos x)^2 - (sin x)^2]/(cos
x)^2

1 - (tan x)^2 = cos 2x/(cos x)^2
(2)

We'll divide (1) by (2) and we'll
get:

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

We'll simplify and we'll
get:

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

We notice that we've obtained, to the left
side, the ratio from the right side, so the given expression is an identity for any
value of x.

1+tanx*tan2x = 1 /
cos 2x