Question

Someone please help me with q6

Answer 1

$\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\\\\\tan3x=\tan(2x+x)=\dfrac{\tan2x+\tan x}{1-\tan2x\tan x}=\dfrac{\tan(x+x)+\tan x}{1-\tan(x+x)\tan x}\\\\=\dfrac{\frac{\tan x+\tan x}{1-\tan x\tan x}+\tan x}{1-\frac{\tan x+\tan x}{1-\tan x\tan x}\tan x}=\dfrac{\frac{2\tan x}{1-\tan^2x}+\tan x}{1-\frac{2\tan x}{1-\tan^2x}\tan x}$

$=\left(\dfrac{2\tan x}{1-\tan^2x}+\dfrac{\tan x(1-\tan^2x)}{1-\tan^2x}\right):\left(\dfrac{1-\tan^2x}{1-\tan^2x}-\dfrac{2\tan^2x}{1-\tan^2x}\right)\\\\=\dfrac{2\tan x+\tan x-\tan^3x}{1-\tan^2x}:\dfrac{1-\tan^2x-2\tan^2x}{1-\tan^2x}\\\\=\dfrac{3\tan x-\tan^3x}{1-\tan^2x}:\dfrac{1-3\tan^2x}{1-\tan^2x}=\dfrac{3\tan x-\tan^3x}{1-\tan^2x}\cdot\dfrac{1-\tan^2x}{1-3\tan^2x}\\\\=\dfrac{3\tan x-\tan^3x}{1}\cdot\dfrac{1}{1-3\tan^2x}=\dfrac{3\tan x-\tan^3x}{1-3\tan^2x}$

$\dfrac{-(\tan^3 x-3\tan x)}{-(3\tan^2x-1)}=\dfrac{\tan^3 x-3\tan x}{3\tan^2x-1}$