Question

prove that:

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

Answer 1

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

L.H.S

$\tan 4x=\tan (2x+2x)$

$=\frac{2\tan 2x}{1-{\tan }^{2}2x}=\frac{2\left(\frac{2\tan x}{1-{\tan }^{2}x}\right)}{1-{\left(\frac{2\tan x}{1-{\tan }^{2}x}\right)}^2}$

$=\frac{4\tan x}{(1-{\tan }^{2}x)}\times \frac{(1-{\tan }^{2}x]{)}^2}{\left[\right(1-{\tan }^{2}x{)}^2-4{\tan }^{2}x]}$

$=\frac{4\tan x[1-{\tan }^{2}x]}{1+{\tan }^{4}x-2{\tan }^{2}x-4{\tan }^{2}x}$

$=\frac{4\tan x[1-{\tan }^{2}x]}{[1-6{\tan }^{2}x+{\tan }^{4}x]}$=R.H.S

LHS=RHS

$\therefore \tan 4x=\frac{4\tan x[1-{\tan }^{2}x]}{[1-6{\tan }^{2}x+{\tan }^{4}x]}$