chứng minh: (sin^2x – cos^2x + cos^4x)/(cos^2x – sin^2x + sin^4x)=tan^2x/cos^2x -tan^2x 18/10/2021 Bởi Adalyn chứng minh: (sin^2x – cos^2x + cos^4x)/(cos^2x – sin^2x + sin^4x)=tan^2x/cos^2x -tan^2x
Giải thích các bước giải: $VT=\dfrac{\sin^2x-\cos^2x+\cos^4x}{\cos^2x-\sin^2x+\sin^4x}\\=\dfrac{\sin^2x-\cos^2x+(\cos^2x)^2}{\cos^2x-\sin^2x+(\sin^2x)^2}\\=\dfrac{\sin^2x+\cos^2x(\cos^2x-1)}{\cos^2x+\sin^2x(\sin^2x-1)}\\=\dfrac{\sin^2x-\cos^2x\sin^2x}{\cos^2x-\sin^2x\cos^2x}\\=\dfrac{\sin^2x(1-\cos^2x)}{\cos^2x(1-\sin^2x)}\\=\dfrac{\sin^2x.\sin^2x}{\cos^2x.\cos^2x}\\=\tan^4x\\VP=\dfrac{\tan^2x}{\cos^2x}-\tan^2x\\=\dfrac{\dfrac{\sin^2x}{\cos^2x}}{\cos^2x}-\dfrac{\sin^2x}{\cos^2x}\\=\dfrac{\sin^2x}{\cos^4x}-\dfrac{\sin^2x}{\cos^2x}\\=\dfrac{\sin^2x-\sin^2x\cos^2x}{\cos^4x}\\=\dfrac{\sin^2x(1-\cos^2x)}{\cos^4x}\\=\dfrac{\sin^2x.\sin^2x}{\cos^4x}\\=\dfrac{\sin^4x}{\cos^4x}\\=\tan^4x=VT\Rightarrow ĐPCM$ Bình luận
Giải thích các bước giải:
$VT=\dfrac{\sin^2x-\cos^2x+\cos^4x}{\cos^2x-\sin^2x+\sin^4x}\\
=\dfrac{\sin^2x-\cos^2x+(\cos^2x)^2}{\cos^2x-\sin^2x+(\sin^2x)^2}\\
=\dfrac{\sin^2x+\cos^2x(\cos^2x-1)}{\cos^2x+\sin^2x(\sin^2x-1)}\\
=\dfrac{\sin^2x-\cos^2x\sin^2x}{\cos^2x-\sin^2x\cos^2x}\\
=\dfrac{\sin^2x(1-\cos^2x)}{\cos^2x(1-\sin^2x)}\\
=\dfrac{\sin^2x.\sin^2x}{\cos^2x.\cos^2x}\\
=\tan^4x\\
VP=\dfrac{\tan^2x}{\cos^2x}-\tan^2x\\
=\dfrac{\dfrac{\sin^2x}{\cos^2x}}{\cos^2x}-\dfrac{\sin^2x}{\cos^2x}\\
=\dfrac{\sin^2x}{\cos^4x}-\dfrac{\sin^2x}{\cos^2x}\\
=\dfrac{\sin^2x-\sin^2x\cos^2x}{\cos^4x}\\
=\dfrac{\sin^2x(1-\cos^2x)}{\cos^4x}\\
=\dfrac{\sin^2x.\sin^2x}{\cos^4x}\\
=\dfrac{\sin^4x}{\cos^4x}\\
=\tan^4x=VT\Rightarrow ĐPCM$
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