Toán chứng minh biểu thức lượng giác sin^2 x-tan^2 x=tan^6 x*(cos^2 x-cot^2 x) 09/07/2021 By Piper chứng minh biểu thức lượng giác sin^2 x-tan^2 x=tan^6 x*(cos^2 x-cot^2 x)
$VP= \tan^6x.(\cos^2x-\cot^2x)$ $=\tan^6x.\dfrac{cos^2x.\sin^2x-\cos^2x}{\sin^2x}$ $=\tan^6x.\dfrac{-\cos^4x}{\sin^2x}$ $=\dfrac{\sin^6x}{\cos^6x}.\dfrac{-\cos^4x}{\sin^2x}$ $=\dfrac{\sin^4x}{-\cos^2x}$ $=\dfrac{\sin^2x.\sin^2x}{-\cos^2x}$ $= \sin^2x.(-\tan^2x)$ $= VT$ Trả lời
Giải thích các bước giải: $\tan^4x(\cos^2x-\cot^2x)$ $=\tan^2x(\tan^2x.\cos^2x-\tan^2x.\cot^2x)$ $=\tan^2x(\sin^2x-1)=-\tan^2x.\cos^2x=-\sin^2x$ $(\sin^2x-\tan^2x).\cot^2x=\sin^2x.\cot^2x-\tan^2x.\cot^2x=\cos^2x-1=-\sin^2x$ $\to (\sin^2x-\tan^2x).\cot^2x=\tan^4x(\cos^2x-\cot^2x)$ $\to (\sin^2x-\tan^2x).\cot^2x.\tan^2x=\tan^6x(\cos^2x-\cot^2x)$ $\to \sin^2x-\tan^2x=\tan^6x(\cos^2x-\cot^2x)$ Trả lời
$VP= \tan^6x.(\cos^2x-\cot^2x)$
$=\tan^6x.\dfrac{cos^2x.\sin^2x-\cos^2x}{\sin^2x}$
$=\tan^6x.\dfrac{-\cos^4x}{\sin^2x}$
$=\dfrac{\sin^6x}{\cos^6x}.\dfrac{-\cos^4x}{\sin^2x}$
$=\dfrac{\sin^4x}{-\cos^2x}$
$=\dfrac{\sin^2x.\sin^2x}{-\cos^2x}$
$= \sin^2x.(-\tan^2x)$
$= VT$
Giải thích các bước giải:
$\tan^4x(\cos^2x-\cot^2x)$
$=\tan^2x(\tan^2x.\cos^2x-\tan^2x.\cot^2x)$
$=\tan^2x(\sin^2x-1)=-\tan^2x.\cos^2x=-\sin^2x$
$(\sin^2x-\tan^2x).\cot^2x=\sin^2x.\cot^2x-\tan^2x.\cot^2x=\cos^2x-1=-\sin^2x$
$\to (\sin^2x-\tan^2x).\cot^2x=\tan^4x(\cos^2x-\cot^2x)$
$\to (\sin^2x-\tan^2x).\cot^2x.\tan^2x=\tan^6x(\cos^2x-\cot^2x)$
$\to \sin^2x-\tan^2x=\tan^6x(\cos^2x-\cot^2x)$