Question

14 БАЛЛОВ ! Решите пж первый пример

Answer 1

$\frac{cos^2(\frac{\pi}{3}+a)}{tg^2(\frac{\pi}{6}-a)}+sin^2(\frac{\pi}{3}+a)\cdot tg^2(\frac{\pi}{6}-a)=$

Для удобства  записи обозначим углы

 $x=\frac{\pi}{3}+a\; ;\; \; y=\frac{\pi}{6}-a\; \; \Rightarrow \; \; x+y=\frac{\pi}{2}\; ;\; \; x-y=\frac{\pi}{6}+2a\; .$

$=\frac{cos^2x}{tg^2y}+sin^2x\cdot tg^2y=\frac{cos^2x\cdot cos^2y}{sin^2y}+\frac{sin^2x\cdot sin^2y}{cos^2y}=\\\\=\frac{(cosx\cdot cosy)^2}{sin^2y}+\frac{(sinx\cdot siny)^2}{cos^2y}=\\\\=\frac{1/4\cdot (cos(x-y)+cos(x+y))^2}{(1-cos2y)/2}+\frac{1/4\cdot (cos(x-y)-cos(x+y))^2}{1/2(1+cos2y)}=\\\\=\Big [\, cos(x+y)=cos\frac{\pi}{2}=0\, \Big ]=\\\\=\frac{1}{2}\cdot \frac{cos^2(x-y)}{1-cos2y}+\frac{1}{2}\cdot \frac{cos^2(x-y)}{1+cos2y}=\\\\=\frac{1}{2}\cdot \frac{cos^2(x-y)\cdot (1+cos2y)+cos^2(x-y)\cdot (1-cos2y)}{1-cos^22y}=$

$=\frac{1}{2}\cdot \frac{cos^2(x-y)+cos^2(x-y)\cdot cos2y+cos^2(x-y)-cos^2(x-y)\cdot cos2y}{1-cos^22y}=\\\\=\frac{1}{2}\cdot \frac{2\cdot cos^2(x-y)}{1-cos^22y}=\frac{cos^2(x-y)}{sin^22y}=\Big [\, 2y=2(\frac{\pi}{6}-a)=\frac{\pi}{3}-2a\, \Big ]=\\\\=\frac{cos^2(\frac{\pi}{6}+2a)}{sin^2(\frac{\pi}{3}-2a)}=\Big [\, sin(\frac{\pi}{3}-2a)=sin(\frac{\pi}{2}-(\frac{\pi}{6}+2a))=cos(\frac{\pi}{6}+2a)\; ,\\\\tak\; kak\; \; \frac{\pi}{3}-2a=\frac{\pi}{2}-t\; \; \to \; \; t=\frac{\pi}{2}-(\frac{\pi}{3}-2a)=\frac{\pi}{6}+2a\, \Big ]=$

$=\frac{cos^2(\frac{\pi}{6}+2a)}{cos^2(\frac{\pi}{6}+2a)}=1\; ,\\\\1=1\; .$