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Question

CM đẳng thứca) cos4α - sin4α = 2cos2α - 1b) $\dfrac{cos^2\alpha+tan^2\alpha-1}{sin^2\alpha}=tan^2\alpha$

Hquynh · Accepted Answer

$a,cos^4a-sin^4a=2cos^2a-1\
VT=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)\
=cos^2a-sin^2a\
=cos2a=2cos^2a-1$$b,VT=\dfrac{cos^2a+\dfrac{sin^2a}{cos^2a}-1}{sin^2a}\
=\dfrac{\dfrac{cos^4a+sin^2a-cos^2a}{cos^2a}}{sin^2a}\
=\dfrac{\dfrac{cos^4a+\left(1-cos^2a\right)-cos^2a}{cos^2a}}{sin^2a}\
=\dfrac{\dfrac{cos^4a+1-2cos^2a}{cos^2a}}{sin^2a}\
=\dfrac{\dfrac{\left(1-cos^2a\right)^2}{cos^2a}}{sin^2a}\
=\dfrac{sin^4a}{cos^2a}:sin^2a\
=\dfrac{sin^4a}{cos^2a}\times\dfrac{1}{sin^2a}\
=\dfrac{sin^2a}{cos^2a}=tan^2a$Câu trả lời: $a,cos^4a-sin^4a=2cos^2a-1\
VT=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)\
=cos^2a-sin^2a\
=cos2a=2cos^2a-1$$b,VT=\dfrac{cos^2a+\dfrac{sin^2a}{cos^2a}-1}{sin^2a}\
=\dfrac{\dfrac{cos^4a+sin^2a-cos^2a}{cos^2a}}{sin^2a}\
=\dfrac{\dfrac{cos^4a+\left(1-cos^2a\right)-cos^2a}{cos^2a}}{sin^2a}\
=\dfrac{\dfrac{cos^4a+1-2cos^2a}{cos^2a}}{sin^2a}\
=\dfrac{\dfrac{\left(1-cos^2a\right)^2}{cos^2a}}{sin^2a}\
=\dfrac{sin^4a}{cos^2a}:sin^2a\
=\dfrac{sin^4a}{cos^2a}\times\dfrac{1}{sin^2a}\
=\dfrac{sin^2a}{cos^2a}=tan^2a$

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