1)(2tg²a*cos²a+2cos²a)*sin²a+3sina=? 2)(1+tg²a)²-1/cos⁴a+3cos²ß+2sin²ß=? - вопрос №2788523122321132 от handogin1999 23.02.2021 12:55

Question

Геометрия: 1)(2tg²a*cos²a+2cos²a)*sin²a+3sina=? 2)(1+tg²a)²-1/cos⁴a+3cos²ß+2sin²ß=? (, решать !

handogin1999 · Answer

(2tg²a*cos²a+2cos²a)*sin²a+3sina=(2sin^2a/cos^2a*cos^2a+2cos^2a)*sin^2a+3sina=
=(2sin^2a+2cos^2a)*sin^2a+3sina=2sin^2a+3sina=sina(2sina+3)
2)(1+tg²a)²-1/cos⁴a+3cos²ß+2sin²ß=1/cos^4a-1/cos^4a+2cos^2b+2sin^2b+cos^2b=
=2+cos^2b