Question

заранее спасибо ...........

Answer 1

$\frac{1-2sin^2 \alpha }{2tg(\frac{\pi}{4}- \alpha )sin^2(\frac{\pi}{4}+ \alpha )}=\\\\=[\, sin \beta =cos(\frac{\pi}{2}- \beta )\; \; \to \\\\sin(\frac{\pi}{4}+ \alpha )=cos(\frac{\pi}{2}-(\frac{\pi}{4}+ \alpha ))=cos(\frac{\pi}{4}- \alpha )\, ]=\\\\=\frac{(sin^2 \alpha +cos^2 \alpha )-2sin^2 \alpha }{2\cdot \frac{sin(\frac{\pi}{4}- \alpha )}{cos(\frac{\pi}{4}- \alpha )}\cdot cos^2(\frac{\pi}{4}- \alpha )}=$

$=\frac{cos^2 \alpha -sin^2 \alpha}{2sin(\frac{\pi}{4}- \alpha )\cdot cos(\frac{\pi}{4}-\alpha)}=$

$=\frac{cos2 \alpha }{sin(2\cdot (\frac{\pi}{4}- \alpha ))}=\frac{cos2 \alpha }{sin(\frac{\pi}{2}-2 \alpha )}=\frac{cos2 \alpha }{cos2 \alpha }=1$