Question

Помогите решить задание из тригонометрии

Answer 1

$\frac{-12}{Sin^{2}131^{0} +Sin^{2}221^{0}} =\frac{-12}{Sin^{2}(90^{0}+41^{0})+Sin^{2} (180^{0}+41^{0})} =\frac{-12}{Cos^{2}41^{0}+Sin^{2}41^{0}} =\\\\=\frac{-12}{1}=\boxed{-12}$

Answer 2

Ответ:

-12

Объяснение:

$\frac{-12}{sin^{2}(90+41)+sin^{2} (180+41) } =\frac{-12}{cos^{2}(41)+sin^{2} (41) } =\frac{-12}{1} =-12$