Question

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Answer 1

Объяснение:

7B.33

a)

$\displaystyle\\\frac{(sin\alpha +cos\alpha )^2-1}{ctg\alpha -sin\alpha* cos\alpha } =\frac{sin^2\alpha +2*sin\alpha *cos\alpha +cos^2\alpha -(sin^2\alpha +cos^2\alpha )}{\frac{cos\alpha }{sin\alpha }-sin\alpha *cos\alpha } =\\\\=\frac{sin^2\alpha +2*sin\alpha *cos\alpha +cos^2\alpha -sin^2\alpha -cos^2\alpha }{\frac{cos\alpha-sin^2\alpha*cos\alpha }{sin\alpha } } =$

$\displaystyle\\=\frac{2*sin\alpha *cos\alpha *sin\alpha }{cos\alpha -sin^2\alpha *cos\alpha }=\frac{2*sin^2\alpha *cos\alpha }{cos\alpha *(1-sin^2\alpha) } =\frac{2*sin^2\alpha }{cos^2\alpha } =2tg^2\alpha .$

б)

$\displaystyle\\\frac{sin\alpha *cos\alpha -tg\alpha }{1-(sin\alpha +cos\alpha )^2}=\frac{sin\alpha *cos\alpha -\frac{sin\alpha }{cos\alpha } }{sin^2\alpha +cos^2\alpha -(sin^2\alpha +2*sin\alpha *cos\alpha +cos^2\alpha )} =\\\\=\frac{sin\alpha *cos^2\alpha -sin\alpha }{cos\alpha *(sin^2\alpha +cos^2\alpha -sin^2\alpha -2*sin\alpha *cosa-cos^2\alpha )} =$

$\displaystyle\\=\frac{sin\alpha *(cos^2\alpha -1)}{cos\alpha *(-2*sin\alpha *cos\alpha )} =\frac{-sin\alpha *(1-cos^2\alpha )}{-2*sin\alpha *cos^2\alpha } =\frac{sin^2\alpha }{2*cos^2\alpha } =\frac{1}{2} tg^2\alpha .$

7B.34

a)

$\displaystyle\\\frac{sin3a+sina}{cos3a+cos\alpha } =\frac{2*sin\frac{3a+a}{2}*cos\frac{3a-a}{2} }{2*cos\frac{3a+a}{2}*cos\frac{3a-a}{2} } =\frac{sin2a*cosa}{cos2a*cosa}=\frac{sin2a}{cos2a}=tg2a.$

б)

$\displaystyle\\\frac{sin72^0-sin18^0}{cos72^0-cos18^0} =\frac{2*sin\frac{72^0-18^0}{2}*cos\frac{72^0+18^0}{2} }{-2*sin\frac{72^0-18^0}{2}*sin\frac{72^0+18^0}{2} } =\frac{cos45^0}{-sin45^0} =-ctg45^0=-1.$