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| 1. |
4tanx(1-tan2x)/1-6tan2x+tan4x |
| Answer» The question is : Prove that tan 4x=\xa04 tanx(1-tan2x)/1-6tan2x+tan4x\xa0L.H.S.tan 4x = tan 2(2x)[We know that tan 2x = 2 tan x / 1 - tan\xa02\xa0x]= 2 tan 2x / 1 - tan2\xa0(2x)[now putting tan 2x = 2 tan x / 1 - tan2x]= 2[2 tan x/1-tan2x] / 1 - [2 tan x / 1 - tan2x]\xa02=[4 tan x / 1 - tan\xa02\xa0x] / [1 - 4 tan\xa02\xa0x / (1 - tan2\xa0x)2]=[4 tan x / 1 - tan\xa02\xa0x] / [ (1- tan\xa02\xa0x)2\xa0- 4 tan\xa02\xa0x / (1 - tan2\xa0x)2]= 4 tan x (1 - tan\xa02\xa0x) / (1- tan\xa02\xa0x)2\xa0- 4 tan\xa02\xa0x= 4 tan x (1 - tan\xa02\xa0x) / 1 - 2 tan2\xa0x +tan\xa04\xa0x - 4tan2\xa0x= 4 tan x (1 - tan\xa02\xa0x) / 1 - 6 tan\xa02\xa0x + tan\xa04\xa0x | |