To prove cos2x(1+tanx*tan2x) = 1

We know that tanx =sinx/cosx.

sin2x = 2sinxcosx.

cos2x = cos^x-sin^2x.

Using the above identities we simplify the Left side:

cos2x(1+ (sinx/cosx )(sin2x/cos2x)

= cos2x +(sinx /cosx)(sin2x)

=cos2x+(sinx/cosx)(2sinx*cosx)

= cos2x +2sin^2x

= cos^2x-sin^2x+2sin^2, as cos2x = cos^2x-sin^2x.

=cos^2x +sin^2x

= 1.

Therefore cos2x(1+tanxtan2x ) = 1 is demonstrated.

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**