Question

докажите тождества
а)
$1 + sin ^{4} alpha - cos ^{4} alpha =2sin ^{2} alpha$
б)
$sin^ {2} 6 alpha cos ^{2}6 alpha (tg ^{2} 6 alpha + ctg ^{2} 6 alpha + 2) = 1$
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Answer 1

$a);1+sin^4alpha-cos^4alpha=1+(sin^2alpha-cos^2alpha)(sin^2alpha+cos^2alpha)=\=1+(sin^2alpha-cos^2alpha)cdot1=1+sin^2alpha-cos^2alpha=sin^2alpha+sin^2alpha=2sin^2alpha$

$b);sin^26alphacos^26alphaleft(tg^26alpha+ctg^26alpha+2right)=\=sin^26alphacos^26alphaleft(frac{sin^26alpha}{cos^26alpha}+frac{cos^26alpha}{sin^26alpha}+2right)=\=sin^26alphacos^26alphaleft(frac{sin^46alpha+cos^46alpha}{sin^26alphacos^26alpha}+2right)=sin^46alpha+cos^46alpha+2sin^26alphacos^26alpha=\=(sin^26alpha+cos^26alpha)^2=1^2=1$