Question

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Answer 1

Ответ:

Объяснение:

1.

$1. =sin^{2} \alpha \\2.=-(1-sin^{2} \alpha)=-cos^{2}\alpha \\3. =cos^{2} \alpha+cos^{2} \alpha=2cos^{2} \alpha \\4. =sin^{2} \alpha+cos^{2} \alpha+cos^{2} \alpha-1=1+cos^{2} \alpha-1=cos^{2} \alpha \\5. =1-sin^{2} \alpha=cos^{2} \alpha \\6. =cos^{2} \alpha-1=-(1-cos^{2} \alpha)=-sin^{2} \alpha\\7. =cos^{2} \alpha-cos^{2} \alpha=0\\8. =cos^{2} \alpha-(1-sin^{2} \alpha-sin^{2} \alpha)=cos^{2} \alpha-(cos^{2} \alpha-sin^{2} \alpha)=sin^{2} \alpha$

$9.=sin^{2} \alpha*cos^{2} \alpha*\frac{sin^{2} \alpha}{cos^{2} \alpha} =sin^{2} \alpha\\10. =sin^{2} \alpha*cos^{2} \alpha*\frac{cos^{2} \alpha}{sin^{2} \alpha}-1 =cos^{2} \alpha -1=-(1-sin^{2} \alpha)=-cos^{2} \alpha\\11. =1+tg^{2} \alpha =\frac{1}{cos^{2} \alpha} \\12. =1+ctg^{2} \alpha =\frac{1}{sin^{2} \alpha}$

$13.=\frac{1}{cos\alpha}-\frac{sin\alpha }{cos\alpha}+tg\alpha =\frac{1}{cos\alpha } \\14. =\frac{1-cos\alpha }{1-cos^{2} \alpha }+\frac{1+cos\alpha }{1cos^{2}\alpha}=\frac{1}{sin^{2} \alpha}$

2.

$1.=\frac{1+2cos\beta sin\beta }{sin^{2}\beta+ 2cos\beta sin\beta+cos^{2}\beta } =\frac{1+2cos\beta sin\beta}{1+2cos\beta sin\beta} =1\\2.=\frac{sin^{2} \beta+(1-sin^{2} \beta)}{sin^{2} \beta} =\frac{2sin^{2} \beta}{sin^{2} \beta} =2\\3.= \frac{1}{\frac{1}{cos^{2} \beta}} +\frac{1}{\frac{1}{sin^{2} \beta}}=cos^{2} \beta+sin^{2} \beta=1\\4.=(\frac{1}{cos \beta}+tg\beta )(\frac{1}{cos \beta}-tg\beta )=\frac{1}{cos^{2} \beta} -tg^{2} \beta =\frac{1-sin^{2}\beta }{cos^{2}\beta } =1$

$5.=2sin\beta cos\beta -2sin\beta cos\beta=0\\6.=(sin^{2} \beta +cos^{2} \beta)^{2} =1\\7.=\frac{2-(sin^{2} \beta+cos^{2} \beta)}{3sin^{2} \beta+3cos^{2} \beta)} =\frac{1}{3} \\8.=\frac{(sin^{2} \beta+cos^{2} \beta)(sin^{2} \beta-cos^{2} \beta)}{(in^{2} \beta-cos^{2} \beta} =1$

3.

$1.(sin^{2} \alpha -sin^{2}\beta )-( cos^{2}\beta -cos^{2}\alpha )=0\\sin^{2} \alpha -sin^{2}\beta-cos^{2}\beta +cos^{2}\alpha=0\\1-(sin^{2}\beta+cos^{2}\beta )=0\\1-1=0\\0=0\\\\2.\frac{cos^{2}\alpha}{sin^{2}\alpha} -\frac{cos^{2}\alpha sin^{2}\alpha}{sin^{2}\alpha} =ctg^{2}\alpha cos^{2}\alpha \\\frac{cos^{2}\alpha(1-sin^{2}\alpha)}{sin^{2}\alpha} =ctg^{2}\alpha cos^{2}\alpha\\ctg^{2}\alpha cos^{2}\alpha=ctg^{2}\alpha cos^{2}\alpha$