2) Найдите tg α, если sin α = -5√41/41, α ∈ (π; 3π/2).

В III четверти \(\cos\alpha<0\):

\[\cos\alpha=-\sqrt{1-\frac{25\cdot41}{41^2}}=-\sqrt{\frac{16}{41}}=-\frac{4\sqrt{41}}{41}.\]

\[ g\alpha=\frac{-5\sqrt{41}/41}{-4\sqrt{41}/41}=\frac54.\]

Ответ: \( g\alpha=\frac54\).