(sin^2 35- cos^2 145)/cos100 *cos 350 с подробным - вопрос №80241892352145100350 от krutikovaelizaozax2f 20.09.2021 23:41

Question

Алгебра: (sin^2 35- cos^2 145)/cos100 *cos 350 с подробным !

krutikovaelizaozax2f · Answer

(Sin²35 - Cos²145) / Cos100*Cos350 = [Sin²35 - Cos²(180 - 35)] / Cos(90 +  +10)*Cos(360 - 10) = (Sin²35 - Cos²35) / (- Sin10)*Cos10 = = - Cos 20 / (- Sin10Cos10) = 2Cos20 / Sin20 = 2 ctg20...