Question

А) tg^2a + ctg^2a, если tg a - ctg a= - 3 б) 6sina + 5cosa/4sina - 3cosa, если tga= 3

Answer 1

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

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

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.