Question

Упростите выражение а) б)

Answer 1

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

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