СРОЧНО! ПОМОГИТЕ, ПОЖАЛУЙСТА! ДАЮ 65 БАЛЛОВ!

Ответ:

Пошаговое объяснение:

1. a³-b³=(a-b)*(a²+ab+b²)

sin³19°-cos³19°=(sin19°-cos19°)*(sin²19°+sin19° * c0s19°+cos²19°)=(sin19°-cos18°)*(1+sin19° *cos19°)

2. sin²57°+sin²33°=sin57°+sin²(90°-57°)=sin²57°+cos²57°=1

3. $tg19^{0}+ctg19^{0}=\frac{sin19^{0}}{cos19^{0}}+\frac{cos19^{0}}{sin19^{0}}=\frac{sin^{2}19^{0}+cos^{2}19^{0}}{sin19^{0}*cos19^{0}}$

=1/(sin19° *cos19°)

4. $\frac{sin^{2}57^{0}+sin^{2}33^{0}}{tg19^{0}+ctg19^{0}}=1:\frac{1}{sin19^{0}*cos19^{0}}=sin19^{0}*cos19^{0}$

5. $\frac{sin^{3}19^{0}-cos^{3}19^{0}}{sin19^{0}-cos19^{0}} -\frac{sin^{2}19^{0}+sin^{2}33^{0}}{tg19^{0}+ctg19^{0}}=\frac{(sin19^{0}-cos19^{0}*(1+sin19^{0}*cos19^{0}))}{sin19^{0}-cos19^{0}}-$

-sin19° *cos19°=1+sin19° *cos19°-sin18° *c0s19°=1