Question

помогите решить задание на фото

Answer 1

$tg^2 alpha =?\\3sin^2 alpha +8cos^2 alpha =4; |:cos^2 alpha ne 0\\3tg^2 alpha +8=frac{4}{cos^2 alpha }\\Formyla:; ; 1+tg^2 alpha =frac{1}{cos^2 alpha }; ; to ; ; frac{4}{cos^2 alpha }=4(1+tg^2 alpha )\\3tg^2 alpha +8=4+4tg^2 alpha \\tg^2 alpha =8-4=4$

Answer 2

$3sin^2x+8cos^2x=4\3(sin^2x+cos^2x)+5cos^2x=4\3+5cos^2x=4\cos^2x=1/5\cos^2x=1-sin^2x\sin^2x=4/5\tg^2x= frac{sin^2x}{cos^2x}=4$