\(sin^4B-cos^4B+2.cos^2B\\ =\left(sin^2B+cos^2B\right).\left(sin^2B-cos^2B\right)+2cos^2B\\ =1.\left(sin^2B-cos^2B\right)+2cos^2B\\ =sin^2B-cos^2B+2cos^2B\\ =sin^2B+cos^2B\\ =1\)
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