Prove the following: tan 4x = [4tan x(1 - tan²x)] / [1 - 6tan²x + tan⁴x]

Solution:

LHS = tan 4x

= tan 2(2x)

= (2tan 2x) / (1 - tan²2x)     [By double angle formulas, tan 2A = (2tan A) / (1 - tan²A)]

= [2(2tan x) / (1 - tan²x)] / [1 - {(2tan x) / (1 - tan²x)²}]     [Again by the same formula]

= [(4tan x) / (1 - tan²x)] / [1 - (4tan²x) / (1 + tan⁴x - 2tan²x)]     [Since (a - b)² = a² + b² - 2ab]

= [(4tan x) / (1 - tan²x)] / [(1 + tan⁴x - 2tan²x - 4tan²x) / (1 + tan⁴x - 2tan²x)]

= [(4tan x) / (1 - tan²x)] × [(1 + tan⁴x - 2tan²x) / (1 + tan⁴x - 6tan²x)]

= [4tan x(1 - tan²x)²] / [(1 - tan²x) (1 + tan⁴x - 6tan²x)]     [As a² + b² - 2ab = (a - b)²]

= [4tan x(1 - tan²x)] / [1 - 6tan²x + tan⁴x]

= RHS

NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.3 Question 23

Prove the following: tan 4x = [4tan x(1 - tan²x)] / [1 - 6tan²x + tan⁴x]

Summary:

We got, tan 4x = [4tan x(1 - tan²x)] / [1 - 6tan²x + tan⁴x]. Hence Proved.