Question

Sin 2x + tan 2x = 4tan/ 1- tan^4 prove that

Answer 1

$\dfrac{4\tan x}{1-\tan^4x}=\dfrac{4\tan x}{(1-\tan^2x)(1+\tan^2x)}$
$=\dfrac{4\tan x}{(1-(\sec^2x-1)\sec^2x}$
$=\dfrac{4\tan x\cos^2x}{2-\sec^2x}$
$=\dfrac{4\sin x\cos x}{2-\sec^2x}$
$=\dfrac{4\sin x\cos^3x}{2\cos^2x-1}$
$=\dfrac{2\sin2x\cos^2x}{\cos2x}$
$=2\tan2x\cos^2x$
$=2\tan2x\left(\dfrac{1+\cos2x}2\right)$
$=\tan2x(1+\cos2x)$
$=\tan2x+\sin2x$