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

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We have to prove that cos 2x * (1 + tan x*tan 2x) = 1

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

cos 2x * (1 + tan x*tan 2x)

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

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

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

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

=> (cos x)^2*(1 - (tan x)^2)*(1 + (tan x)^2)/(1 - (tan x)^2))

=> (cos x)^2*(1 + (tan x)^2)

=> (cos x)^2*(1 + (sin x)^2/(cos x)^2)

=> (cos x)^2*((cos x)^2 + (sin x)^2)/(cos x)^2

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

=> 1

**This proves that cos 2x * (1 + tan x*tan 2x) = 1**

We'll divide both sides by cos 2x.

1+tanx*tan2x = 1/cos 2x

Now, we'll recall the double angle identity for tan 2x:

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

tan (2x) = 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 as a fraction:

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 thePythagorean identity, we'll get:

(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

**Since LHS=RHS, therefore the given expression is verified for any value of x.**