a) $\frac{1+sin^2x}{1-sin^2x}=1+tan^2x$
b) $\frac{tan^2a-sin^2a}{cot^2a-cos^2a}=tan^6a$

$\frac{tan^2x-sin^2x}{cot^2x-cos^2x}=\frac{\frac{sin^2x}{cos^2x}-sin^2x}{\frac{cos^2x}{sin^2x}-cos^2x}=\frac{sin^2x\frac{1-cos^2x}{cos^2x}}{cos^2x\frac{1-sin^2x}{sin^2x}}=\frac{\frac{sin^4x}{cos^2x}}{\frac{cos^4x}{sin^2x}}=\frac{sin^6x}{cos^6x}=tan^6x$


$DPCM$

$VP(1)=1+\frac{sin^2x}{cos^2x}=\frac{1+sin^2x}{cos^2x}=\frac{1+sin^2x}{1-sin^2x}=VT$

$DPCM$

có vẻ như là bị thừa thì phải. thừa 1 ở phần tử của biểu thức đó. cm kiểu gì cũng k được. –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 28-08-14 10:05 PM
để mk xem –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 28-08-14 10:02 PM
bạn có thể làm lại bài này đc ko? :D –  smix84 28-08-14 09:59 PM
thôi chết. nhầm.hjz –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 28-08-14 09:58 PM
tại sao 1 (sin^2x)/(cos^2x) lại bằng (1 sin^2x)/(cos^2x) hả bạn? mình ko hiểu –  smix84 28-08-14 09:51 PM

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