Prove that    tan4x= (4tanx(1-tan^2x)) / 1-6tan^x+tan^4x

we know tan2Q = 2tanQ / 1-tan2Q

using this formula,

tan4x = tan2.(2x)

=2tan2x / 1-tan22x

=  2(2tanx/1-tan2x)  /  1-(2tanx/{1-tan2x}2)

simplify to get. ,  4tanx(1-tan2x) / 1-6tan2x+tan4x.

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sin26x-sin24x=sin2xsin10x

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sinx26x-sin24x=sin2xsin10x

[sin(6x-4x)]2

sin(6x+4x)sin(6x-4x)

sin10xsin2x

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tan 4x = tan 2(2x) [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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Prove that cos 8 minus sin 8 upon cos 8 + sin True

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This helps

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:-P
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Please find your solution below

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@mohit mishra
hope it will help you
 
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Please find this answer

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Solution to class 11 ncert problem

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Please find this answer

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My dear your answer is here

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And then

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Sorry the another solution was posted by mistake :(
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May it be helpful

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