Question

Даю 20 баллов помогите пожалуйста номер 14

Answer 1

$\frac{2cos^2(16) + 2cos^2(76) - 3}{cos^2(44)} = \frac{cos(32) + cos(152) - 1}{cos^2(44)} = \frac{2cos(92)cos(60) - 1}{cos^2(44)} = \frac{cos(92) - 1}{cos^2(44)} = -\frac{cos(88) + 1}{1 + cos(88)} =\\= -1$