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**