Question

Докажите тождество
$\frac{sin^4a-sin^2a}{sin^3a-cos^2a*sina} = \frac{1}{2} tg2acosa$

Answer 1

$\frac{Sin ^{4} \alpha -Sin ^{2} \alpha}{Sin ^{3} \alpha - Cos ^{2} \alpha Sin \alpha }= \frac{Sin ^{2} \alpha (Sin ^{2} \alpha -1)}{Sin \alpha (Sin ^{2} \alpha -Cos ^{2} \alpha ) }= \frac{Sin \alpha Cos ^{2} \alpha }{Cos2 \alpha }= \frac{2Sin \alpha Cos \alpha Cos \alpha }{2Cos2 \alpha }=$$\frac{Sin2 \alpha Cos \alpha }{2Cos2 \alpha } = \frac{1}{2} tg2 \alpha Cos \alpha \\\\ \frac{1}{2}tg2 \alpha Cos \alpha = \frac{1}{2} tg2 \alpha Cos \alpha$
Тождество доказано