Lượng giác...em ko bít làm ToT

N

nguyenbahiep1

[laTEX]\frac{(1-cosx)(1+cosx)}{(1-sinx)(1+sinx)}.(1-sinx)(1+sinx+sin^2x) = (1-cosx)(1+cosx+cos^2x) \\ \\ TH_1: cosx = 1 \\ \\ (1+cosx)(1+sinx+sin^2x) = (1+sinx)(1+cosx+cos^2x) \\ \\ sin^2x +cosx.sin^2x = cos^2x + sinx.cos^2x \\ \\ (sinx-cosx)(sinx+cosx) + cosx.sinx(sinx-cosx) = 0 \\ \\ TH_2: sinx = cosx \\ \\ TH_3:sinx+cosx + sinx.cosx = 0 \\ \\ sinx+ cosx = t [/laTEX]
 
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