Question

Помогите пожалуйста очень срочно!!!!!Умоляю!!!
а) tg^2a + ctg^2a, если tg a - ctg a= - 3
б) 6sina + 5cosa/4sina - 3cosa, если tga= 3

Answer 1

а)
$tg alpha -ctg alpha =-3 \ frac{sin alpha }{cos alpha }- frac{cos alpha }{sin alpha }=-3 \ \ frac{sin^2 alpha -cos^2 alpha }{sin alpha *cos alpha }=-3 \ \ frac{-(cos^2 alpha -sin^2 alpha )}{ frac{2sin alpha *cos alpha }{2} }=-3 \ \ - frac{2cos 2alpha }{sin 2alpha }=-3 \ \ ctg2 alpha =1.5$

$tg^2 alpha +ctg^2 alpha = frac{sin^2 alpha }{cos^2 alpha }+ frac{cos^2 alpha }{sin^2 alpha }= frac{sin^4 alpha +cos^4 alpha }{sin^2 alpha *cos^2 alpha }= \ \ = frac{(sin^2 alpha +cos^2 alpha )^2-2sin^2 alpha *cos^2 alpha }{ frac{4sin^2 alpha *cos^2 alpha }{4} }= \ \ = frac{4(1-(2sin alpha *cos alpha )( frac{1}{2}*2sin alpha *cos alpha ))}{sin^22 alpha } = \ \ = frac{4(1-0.5sin^22 alpha )}{sin^22 alpha }= frac{4-2sin^22 alpha }{sin^22 alpha }=$
$= frac{4}{sin^22 alpha }-2=4* frac{1}{sin^22 alpha }-2= \ \ =4*(ctg^22 alpha +1)-2=4*(1.5^2+1)-2= \ \ =4*(2.25+1)-2=4*3.25-2= 13-2=11$

Ответ: 11.

б)
$tg alpha =3 \ frac{sin alpha }{cos alpha }=3 \ \ sin alpha =3cos alpha$

$frac{6sin alpha +5cos alpha }{4sin alpha -3cos alpha }= frac{6*3cos alpha +5cos alpha }{4*3cos alpha -3cos alpha }= frac{23cos alpha }{9cos alpha }= frac{23}{9}=2 frac{5}{9}$

Ответ: 2 ⁵/₉.