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

**Solution:**

LHS = tan 4x

= tan 2(2x)

= (2tan 2x) / (1 - tan^{2}2x) [By double angle formulas, tan 2A = (2tan A) / (1 - tan^{2}A)]

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

= [(4tan x) / (1 - tan^{2}x)] / [1 - (4tan^{2}x) / (1 + tan^{4}x - 2tan^{2}x)] [Since (a - b)^{2} = a^{2} + b^{2} - 2ab]

= [(4tan x) / (1 - tan^{2}x)] / [(1 + tan^{4}x - 2tan^{2}x - 4tan^{2}x) / (1 + tan^{4}x - 2tan^{2}x)]

= [(4tan x) / (1 - tan^{2}x)] × [(1 + tan^{4}x - 2tan^{2}x) / (1 + tan^{4}x - 6tan^{2}x)]

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

= [4tan x(1 - tan^{2}x)] / [1 - 6tan^{2}x + tan^{4}x]

= 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 - tan^{2}x)] / [1 - 6tan^{2}x + tan^{4}x]. Hence Proved.

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