Giải thích các bước giải:
$\tan^4x(\cos^2x-\cot^2x)$
$=\tan^2x(\tan^2x.\cos^2x-\tan^2x.\cot^2x)$
$=\tan^2x(\sin^2x-1)=-\tan^2x.\cos^2x=-\sin^2x$
$(\sin^2x-\tan^2x).\cot^2x=\sin^2x.\cot^2x-\tan^2x.\cot^2x=\cos^2x-1=-\sin^2x$
$\to (\sin^2x-\tan^2x).\cot^2x=\tan^4x(\cos^2x-\cot^2x)$
$\to (\sin^2x-\tan^2x).\cot^2x.\tan^2x=\tan^6x(\cos^2x-\cot^2x)$
$\to \sin^2x-\tan^2x=\tan^6x(\cos^2x-\cot^2x)$