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

neela | Student

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.

giorgiana1976 | Student

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