Question

Выражение: sin2a (sin a + cos a) ^2 -1

Answer 1

$\frac{2sinxcosx}{1 + 2sinxcosx-1} = \frac{2sinxcosx}{2sinxcosx}=1$

Answer 2

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