Question

Prove that ;$\tan 4x=\frac{4\tan x(1-{\tan }^{2}x)}{1-6{\tan }^{2}x+{\tan }^{4}x}$.

Answer 1

$\begin{array}{cc}\tan 4x=\frac{4\tan x\left(1-{\tan }^{2}x\right)}{1-6{\tan }^{2}x+{\tan }^{4}x}& \\ =\tan \left(3x+x\right)=\frac{\tan 3x+\tan x}{1-\tan 3x.\tan x}& \left[\tan 3x=\frac{\left(3\tan x-{\tan }^{3}x\right)}{\left(1-3{\tan }^{2}x\right)}\right]\\ =\frac{\left(\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}\right)+\tan x}{1-\left(\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}\right)\tan x}& \\ =\frac{\frac{3\tan x-{\tan }^{3}x+\tan x-3{\tan }^{3}x}{\left(1-3{\tan }^{2}x\right)}}{\frac{1-3{\tan }^{2}x-3{\tan }^{2}x+{\tan }^{4}x}{\left(1-3{\tan }^{2}x\right)}}& \\ =\tan 4x=\frac{4\tan x-4{\tan }^{3}x}{1-\left(6{\tan }^{2}x-{\tan }^{4}x\right)}& \end{array}$