Question

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

Answer 1

$TakingLHS$

$Weknowthat$

$tan2x=\frac{2tanx}{1-{tan}^{2}x}$

$Replacingxwithtanx$

$tan(2\ast 2x)=\frac{2tan2x}{1-{tan}^{2}2x}$

$tan\left(4x\right)=\frac{2tan2x}{1-{tan}^{2}2x}$

$Usingtan2x=\frac{2tanx}{1-{tan}^{2}x}$

$tan4x=\frac{2\left(\frac{2tanx}{1-{tan}^{2}x}\right)}{{1-\left(\frac{2tanx}{1-{tan}^{2}x}\right)}^2}$

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

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

$=\frac{4tanx}{{1-tan}^{2}x}\ast \frac{{(1-tan}^{2}x)2}{{{(1-tan}^{2}x)}^2-4{tan}^{2}x}$

$=\frac{4tanx}{1}\ast \frac{{(1-tan}^{2}x)}{{{(1-tan}^{2}x)}^2-4{tan}^{2}x}$

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

$Using{(a-b)}^2=a^2+b^2-2ab$

$=\frac{4tanx(1-{tan}^{2}x)}{\left(\right(1^2)+{\left({tan}^{2}x\right)}^2-2\ast 1\ast {tan}^{2}x)-4{tan}^{2}x}$

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

$=RHS$

$Henceproved$