Question

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

Answer 1

$tan4x=tan2\left(2x\right)=\frac{2tan2x}{(1-tan²2x)}=\frac{2\ast \frac{2tanx}{(1-tan²x)}}{\frac{1-4tan²x}{(1-tan²x)²}}$

At this step, we can cancel one $tan²x$

=$\frac{4tanx(1-tan²x)}{(ta{n}^{4}x-2tan²x+1)-4tan²x}$

$=\frac{4tanx(1-tan²x)}{(1-6tan²x+ta{n}^{4}x)}$

Therefore comparing the two equation we have $M=4$