4+cosa+tg^2a если cosa=12
Cosα =1/2 .
---
(4+cosα +tq²α) - ?
4+cosα +tq²α =4+cosα +tq²α =4+cosα +(1-cos²α)/cos²α =
4+1/2 +(1-(1/2)²)/(1/2)² =4+1/2 +(1-1/4)/(1/4) =4+1/2+(3/4)/(1/4)=4+1/2 +3 =7,5.
Оцени ответ
Вход
Регистрация
Задать вопрос
Cosα =1/2 .
---
(4+cosα +tq²α) - ?
4+cosα +tq²α =4+cosα +tq²α =4+cosα +(1-cos²α)/cos²α =
4+1/2 +(1-(1/2)²)/(1/2)² =4+1/2 +(1-1/4)/(1/4) =4+1/2+(3/4)/(1/4)=4+1/2 +3 =7,5.