Question

(2sin(a) - sin(2a))/(2cos(a)-cos(2a)-1)

Answer 1

Ответ:

$tg\alpha$

Объяснение:

$\dfrac{2sin\alpha-sin2\alpha}{2cos\alpha-cos2\alpha-1}=\dfrac{2sin\alpha-2sin\alpha cos\alpha}{2cos\alpha-(2cos^{2}\alpha-1)-1}=\dfrac{2sin\alpha(1-cos\alpha)}{2cos\alpha-2cos^{2}\alpha+1-1}=$

$=\dfrac{2sin\alpha(1-cos\alpha)}{2cos\alpha(1-cos\alpha)}=\dfrac{2sin\alpha}{2cos\alpha}=\dfrac{sin\alpha}{cos\alpha}=tg\alpha;$