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

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

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

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

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