Нужен номер 5 и 6 . Подробное решение. Created by art2131. algebra-ru

Нужен номер 5 и 6 . Подробное решение...

art2131 @art2131

September 2021 1 20 Report

Нужен номер 5 и 6 . Подробное решение

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kirichekov

Verified answer

$tg \alpha -ctg \alpha =-3. tg ^{2} \alpha +ctg ^{2} \alpha =?

(tg \alpha -ctg \alpha ) ^{2} =(-3)^{2} 

tg ^{2} \alpha -2tg \alpha *ctg \alpha +ctg ^{2} \alpha =9$

$tg ^{2} \alpha -2+ctg ^{2} \alpha =9

tg ^{2} \alpha +ctg ^{2} \alpha =11$

5. b). tgα=3.
$\frac{sin \alpha }{cos \alpha } =3, sin \alpha =3*cos \alpha$

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

6.
$arcsin0-arccos \frac{ \sqrt{2} }{2}- \frac{arctg \frac{ \sqrt{3} }{3} }{arcctg \sqrt{3} } =0- \frac{ \pi }{4} - \frac{ \frac{ \pi }{6} }{ \frac{ \pi }{6} } =- \frac{ \pi }{4} -1$

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