A=(1-sin4x-cos4x)/(1+cos4x+sin4x). Rút gọn

Giải thích các bước giải:

Ta có:

$A=\dfrac{1-\sin4x-\cos4x}{1+\cos4x+\sin4x}$

$\to A=\dfrac{1-\sin4x-(1-2\sin^22x)}{1+(2\cos^22x-1)+\sin4x}$

$\to A=\dfrac{-\sin4x+2\sin^22x}{2\cos^22x+\sin4x}$

$\to A=\dfrac{-2\sin2x\cos2x+2\sin^22x}{2\cos^22x+2\sin2x\cos2x}$

$\to A=\dfrac{2\sin2x(\sin2x-\cos2x)}{2\cos2x(\sin2x+\cos2x)}$

$\to A=\dfrac{\sin2x(\sin2x-\cos2x)}{\cos2x(\sin2x+\cos2x)}$

$\to A=\dfrac{\dfrac{\sin2x}{\cos2x}-1}{1+\dfrac{\cos2x}{\sin2x}}$

$\to A=\dfrac{\tan2x-1}{1+\dfrac1{\tan2x}}$

$\to A=\dfrac{\tan2x-1}{1+\cot2x}$