prove that tan4x= tanx(1-tan^2x)/1-6tan^2x+tan^4x
Answers
Step-by-step explanation:
We know that tan 2x = 2 tan x / 1 - tan 2 x]
= 2 tan 2x / 1 - tan2 (2x)
[now putting tan 2x = 2 tan x / 1 - tan2x]
= 2[2 tan x/1-tan2x] / 1 - [2 tan x / 1 - tan2x] 2
=[4 tan x / 1 - tan 2 x] / [1 - 4 tan 2 x / (1 - tan2 x)2]
=[4 tan x / 1 - tan 2 x] / [ (1- tan 2 x)2 - 4 tan 2 x / (1 - tan2 x)2]
= 4 tan x (1 - tan 2 x) / (1- tan 2 x)2 - 4 tan 2 x
= 4 tan x (1 - tan 2 x) / 1 - 2 tan2 x +tan 4 x - 4tan2 x
= [4 tan x (1 - tan 2 x) / 1 - 6 tan 2 x + tan 4 x].
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Step-by-step explanation:
LHS = tan4x
= tan(2x + 2x)
use, the formula,
tan(A + B) = (tanA+tanB)/(1-tanA.tanB)
= (tan2x + tan2x)/(1-tan2x.tan2x)
=2tan2x/(1-tan²2x)
again, use the formula,
tan2A = 2tanA/(1-tan²A)
= 2{2tanx/(1-tan²x)}/[1-{2tanx/(1-tan²x)}²]
=4tanx.(1-tan²x)²/(1-tan²x)(1+tan⁴x-2tan²x-4tan²x)
=4tanx.(1-tan²x)/(1-6tan²x+tan⁴x) = RHS