CMR: `\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}=1+tan^2x` 29/08/2021 Bởi Emery CMR: `\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}=1+tan^2x`
$\frac{cos²x – sin²x}{sin³x+cos⁴x-sin²x}$ = $\frac{\frac{1}{cos²x} tan²x.\frac{1}{cos²x}}{tan⁴x-1 tan²x\frac{1}{cos²x} }$ =$\frac{1+tan²x-tan²x (1+tan²x)}{tan⁴x+1-tan²x(1+tan²x)}$ =$\frac{(1-tan²x)(1-tan²x)}{1-tan²x}$ = $\frac{1-tan⁴x}{1-tan²x}$ = 1+ tan²x Mk sửa lại r cho mk ctlhn nha *** Bình luận
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$\frac{cos²x – sin²x}{sin³x+cos⁴x-sin²x}$
= $\frac{\frac{1}{cos²x} tan²x.\frac{1}{cos²x}}{tan⁴x-1 tan²x\frac{1}{cos²x} }$
=$\frac{1+tan²x-tan²x (1+tan²x)}{tan⁴x+1-tan²x(1+tan²x)}$
=$\frac{(1-tan²x)(1-tan²x)}{1-tan²x}$
= $\frac{1-tan⁴x}{1-tan²x}$
= 1+ tan²x
Mk sửa lại r cho mk ctlhn nha ***