Question

Доказать тождество
$frac{tg3 beta }{tg2 beta } -1- frac{tg beta }{tg2 beta } =tg beta tg3 beta$

Answer 1

$beta =b\ frac{tg3b}{tg2b}-1-frac{tgb}{tg2b}=tgb*tg3b\\ tg3b=frac{sin3b}{cos3b}=frac{(4cos^2b-1)sinb}{4cos^3b-3cosb}\\ tg2b=frac{sin2b}{cos2b}=frac{2sinb*cosb}{2cos^2b-1}\\ frac{tg3b}{tg2b} = frac{8cos^4b-6cos^2b+1}{8cos^4b-6cos^2b}\\ frac{tgb}{tg2b}=frac{2cos^2b-1}{2cos^2b}\\ tgb*tg3b=frac{ 3sin^2b-4sin^4b}{4cos^4b-3cos^2b}\\ frac{8cos^4b-6cos^2b+1}{8cos^4b-6cos^2b}-frac{2cos^2b-1}{2cos^2b}-1=frac{3sin^2b-4sin^4b}{4cos^4b-3cos^2b}$
то есть  надо теперь доказать это выражение   
 $frac{8cos^4b-6cos^2b+1}{8cos^4b-6cos^2b}-frac{2cos^2b-1}{2cos^2b}-1=frac{3sin^2b-4sin^4b}{4cos^4b-3cos^2b}\\ frac{8cos^4b-6cos^2b+1}{2cos^2b(4cos^2b-3)}-frac{(2cos^2b-1)(4cos^2b-3)}{2cos^2b(4cos^2b-3)}$
$frac{8cos^4b-6cos^2b+1}{2cos^2b(4cos^2b-3)}-frac{(2cos^2b-1)(4cos^2b-3)}{2cos^2b(4cos^2b-3)}-\ frac{2cos^2b(4cos^2b-3)}{2cos^2b(4cos^2b-3)}=frac{3sin^2b-4sin^4b}{4cos^4b-3cos^2b}\\ frac{8cos^4b-6cos^2b+1-(2cos^2b-1)(4cos^2b-3)-2cos^2b(4cos^2b-3)}{2cos^2b(4cos^2b-3)}=\\ frac{6sin^2b-8sin^4b}{8cos^4b-6cos^2b}=frac{3sin^2b-4sin^4b}{4cos^4b-3cos^2b}$ 
то есть верно