Ta có: \(\frac{1}{cos^2x}=\frac{sin^2x+cos^2x}{cos^2x}=\frac{\frac{sin^2x}{cos^2x}+\frac{cos^2x}{cos^2x}}{\frac{cos^2x}{cos^2x}}=tan^2x+1\)
Vậy \(\frac{1}{cos^2x}=1+tan^2x\)
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