Question

Someone please help me with q6

Answer 1

`\tan(x+y)=(\tan x+\tan y)/(1-\tan x\tan y)\\\\\tan3x=\tan(2x+x)=(\tan2x+\tan x)/(1-\tan2x\tan x)=(\tan(x+x)+\tan x)/(1-\tan(x+x)\tan x)\\\\=((\tan x+\tan x)/(1-\tan x\tan x)+\tan x)/(1-(\tan x+\tan x)/(1-\tan x\tan x)\tan x)=((2\tan x)/(1-\tan^2x)+\tan x)/(1-(2\tan x)/(1-\tan^2x)\tan x)`

`=\left((2\tan x)/(1-\tan^2x)+(\tan x(1-\tan^2x))/(1-\tan^2x)\right):\left((1-\tan^2x)/(1-\tan^2x)-(2\tan^2x)/(1-\tan^2x)\right)\\\\=(2\tan x+\tan x-\tan^3x)/(1-\tan^2x):(1-\tan^2x-2\tan^2x)/(1-\tan^2x)\\\\=(3\tan x-\tan^3x)/(1-\tan^2x):(1-3\tan^2x)/(1-\tan^2x)=(3\tan x-\tan^3x)/(1-\tan^2x)\cdot(1-\tan^2x)/(1-3\tan^2x)\\\\=(3\tan x-\tan^3x)/(1)\cdot(1)/(1-3\tan^2x)=(3\tan x-\tan^3x)/(1-3\tan^2x)`

`(-(\tan^3 x-3\tan x))/(-(3\tan^2x-1))=(\tan^3 x-3\tan x)/(3\tan^2x-1)`