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Marisabel10061999 @Marisabel10061999

July 2022 1 14 Report

упростите выражение а)[tex] \frac{(sin \alpha -cos \alpha ) ^{2}-1 }{ctg \alpha -sin \alpha cos \alpha } [/tex]
б)[tex] \frac{tg ( \pi + \alpha )}{1-ctg ^{2}( \frac{ \pi }{2}+ \alpha ) } * \frac{1-tg ^{2}( \frac{3 \pi }{2} - \alpha ) }{ctg \alpha } [/tex]

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

$=-2tg^2 \alpha;\\\\b)\ \frac{tg(\pi+ \alpha 
)}{1-ctg^2(\frac{\pi}{2}+ \alpha )}\bullet\frac{1-tg^2(\frac{3\pi}{2}- 
\alpha )}{ctg \alpha }=\frac{tg\alpha }{1-tg^2\alpha 
}\bullet\frac{1-ctg^2\alpha }{ctg \alpha }=\frac{tg\alpha 
(1-ctg^2\alpha)}{ctg \alpha (1-tg^2\alpha) }=\\\\=\frac{tg\alpha 
(1-ctg^2\alpha)}{ctg \alpha (1-tg^2\alpha) 
}=\frac{tg\alpha-ctg\alpha}{ctg \alpha-tg\alpha }=\frac{-(ctg \alpha -tg
 \alpha )}{(ctg \alpha -tg \alpha )}=-1.$

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