Question

Помогите, заранее спасибо:
ДОКАЗАТЬ ТОЖДЕСТВО:
(√2cos a - 2cos(Π/4))/(2sin(Π/6+a) - √3sin a = -√2tg a

Answer 1

$\frac{\sqrt2cosa-2cos(\frac{\pi}{4}-a)}{2sin(\frac{\pi}{6}+a)-\sqrt3sina} = \frac{\sqrt2cosa-2(cos\frac{\pi}{4}\cdot cosa+sin\frac{\pi}{4}\cdot sina)}{2(sin\frac{\pi}{6}\cdot cosa+cos\frac{\pi}{6}\cdot sina)-\sqrt3sina} =\\\\= \frac{\sqrt2cosa-\sqrt2cosa-\sqrt2sina}{cosa+\sqrt3sina-\sqrt3sina} = \frac{-\sqrt2sina}{cosa} =-\sqrt2tga\\\\\star \; \; sin \frac{\pi}{4}=cos\frac{\pi}{4} = \frac{\sqrt2}{2}$

Answer 2

(√2сosa-2cosπ/4cosa-2sinπ/4sina)/(2sinπ/6cosa+2cosπ/6sina-√3sina)=
=(√2cosa-2*√2/2*cosa-2*√2/2*sina)/(2*1/2*cosa+2*√3/2sina-√3sina)=
=(√2cosa-√2cosa-√2sina)/(cosa+√3sina-√3sina)=-√2sina/cosa=-√2tga